Show that $\left(\nabla_c\nabla_d-\nabla_d\nabla_c\right)v^a=R_{bcd}^av^b$ for vector field, $v$, and the Riemann curvature tensor, $R_{bcd}^a$ I'm going to be asking about parts of the author's solution to the following question. I feel this is a very important question to understand as it touches on some crucial aspects of the covariant derivative.

Show that (without torsion)
$$\left(\nabla_c\nabla_d-\nabla_d\nabla_c\right)v^a=R_{bcd}^av^b\tag{1}$$
where
$$R_{bcd}^a=\partial_c\Gamma_{bd}^a-\partial_d \Gamma_{bc}^a+\Gamma_{ec}^a\Gamma_{bd}^e-\Gamma_{ed}^a\Gamma_{bc}^e$$
is the Riemann curvature tensor. Hint: determine $\nabla_c\nabla_dv^a$.


I'm trying to understand the following solution to the question above:

$$\nabla_c\nabla_dv^a=\partial_c\nabla_dv^a-\Gamma_{dc}^b\nabla_b v^a+\Gamma_{bc}^a\nabla_d v^b\tag{a}$$
$$=\partial_c\left(\partial_dv^a+\Gamma_{bd}^av^b\right)-\Gamma_{dc}^b\left(\partial_bv^a+\Gamma_{eb}^av^e\right)+\Gamma_{bc}^a\left(\partial_dv^b+\Gamma_{ed}^bv^e\right)\tag{b}$$
$$=\left(\partial_c\Gamma_{bd}^a\right)v^b+\Gamma_{bc}^a\Gamma_{ed}^bv^e + \text{contribution symmetric in $c$ and $d$ indices}\tag{c}$$
$$=\left(\partial_c\Gamma_{bd}^a\right)v^b+\Gamma_{ec}^a\Gamma_{bd}^ev^b + \text{contribution symmetric in $c$ and $d$ indices}\tag{d}$$
on exchanging dummy $b$ and $e$ indices in the second term. Subtracting $\nabla_d\nabla_cv^a$ gives the stated result.


In order for me to make sense of the author's solution, I need to understand each of the four equalities, $(\mathrm{a})-(\mathrm{d})$, in turn.
So starting with $(\mathrm{a})$, for a contravariant vector field $v^a(x)$, the covariant derivative, $\nabla_{\text{cov}}$ is defined through
$$\nabla_{\text{cov}}v^a=\partial_{\text{cov}}v^a+\Gamma_{b,\,{\text{cov}}}^av^b\tag{2}$$
Similarly, for a covariant vector field $v_b(x)$, the covariant derivative, $\nabla_{\text{cov}}$ is defined through
$$\nabla_{\text{cov}}v_b=\partial_{\text{cov}}v_b-\Gamma_{b,\,{\text{cov}}}^av_a\tag{3}$$
[Please note that I have purposely used '$\mathrm{cov}$' to represent the index for the covariant derivative, this was done to avoid writing say, '$\nabla_c$', which would cause confusion with the notation that follows].
For a tensor field of type $(1,1)$, $T_b^a$, the covariant derivative of this tensor field, $\nabla_c$ is
$$\nabla_c T_b^a=\partial_c T_{b}^a+\Gamma_{cd}^a T_{b}^d-\Gamma_{bc}^e T_{e}^a \tag{4}$$
Since the gradient of a vector field is a tensor (field), let the tensor (field) of type $(1,1)$ be $T_d^a=\nabla_d v^a$. Now, using the hint at the end of the question and eqn $(4)$,
$$\begin{align}\nabla_c\left(\nabla_dv^a\right)&=\nabla_cT_d^a\\&=\partial_cT_d^a+\Gamma_{bc}^aT_d^b-\Gamma_{cd}^e T_e^a\\&=\partial_c\nabla_dv^a+\Gamma_{bc}^a\nabla_dv^b-\Gamma_{cd}^e \nabla_ev^a\tag{5}\end{align}$$
and this is the equivalent of $(\mathrm{a})$ in the author's solution.
The next equality, $(\mathrm{b})$, can be found by direct substitution of eqn $(2)$ into $(5)$.
To try to understand the last two equalities, $(\mathrm{c})-(\mathrm{d})$, I multiply out eqn $(\mathrm{b})$:
$$\begin{align}\nabla_c\nabla_dv^a &=\partial_c\left(\partial_dv^a+\Gamma_{bd}^av^b\right)-\Gamma_{dc}^b\left(\partial_bv^a+\Gamma_{eb}^av^e\right)+\Gamma_{bc}^a\left(\partial_dv^b+\Gamma_{ed}^bv^e\right)\\&=\partial_c\partial_dv^a+\partial_c\Gamma_{bd}^av^b-\Gamma_{dc}^b\partial_bv^a-\Gamma_{dc}^b\Gamma_{eb}^av^e+\Gamma_{bc}^a\partial_dv^b+\Gamma_{bc}^a\Gamma_{ed}^bv^e\tag{6}\\& \stackrel{\color{red}{?}}{=}\left(\partial_c\Gamma_{bd}^a\right)v^b+\Gamma_{bc}^a\Gamma_{ed}^bv^e+ \text{contribution symmetric in $c$ and $d$ indices}\tag{7}\end{align}$$
But the final equality (marked with a red question mark above the equals sign) can only be true if out of the six terms in eqn $(6)$ we have
$$\partial_c\partial_dv^a=-\Gamma_{dc}^b\partial_bv^a=\Gamma_{bc}^a\partial_dv^b=0\tag{8}$$
and therefore I identify $$-\Gamma_{dc}^b\Gamma_{eb}^av^e\tag{9}$$ as the contribution symmetric in $c$ and $d$ indices in $(7)$.
I'll reserve asking about part $(\mathrm{d})$ for now until I have understood part $(\mathrm{c})$.

So my questions about eqn $(\mathrm{c})$ are:

*

*What justifies those particular three terms from eqn $(6)$ being equal to zero in $(8)$, and why are the others non-zero?

*Is $(9)$ the unidentified symmetric contribution in the $c$ and $d$ indices eluded to $(7)$?


Here I embed images of the question and its' solution in case I made any typos while typesetting this post:



and here is the solution:




Update:
Even though a good answer has already been provided the part I still can't understand is the resulting expression $(\mathrm{c})$: $$\nabla_c\nabla_dv^a=\left(\partial_c\Gamma_{bd}^a\right)v^b+\Gamma_{bc}^a\Gamma_{ed}^bv^e + \text{contribution symmetric in $c$ and $d$ indices}$$
I understand how $(\mathrm{d})$ was obtained from $(\mathrm{c})$ but since my question revolves almost entirely about how to get from eqn $(\mathrm{b})$ to $(\mathrm{c})$, for the purposes of this question and to stay consistent with the answer given by @ContraKinta I will use eqn $(\mathrm{c})$.
I think the line "contribution symmetric in $c$ and $d$ indices" is the part causing me the most confusion. For instance, if I were to explicitly write down these 'symmetric contributions' when indices $c$ is switched with $d$ and vice versa then I think the resulting expression should be:
$$\nabla_c\nabla_dv^a=\left(\partial_c\Gamma_{bd}^a\right)v^b+\Gamma_{bc}^a\Gamma_{ed}^bv^e+\left(\partial_{\color{blue}{d}}\Gamma_{b\color{red}{c}}^a\right)v^b+\Gamma_{b \color{blue}{d}}^a\Gamma_{e\color{red}{c}}^bv^e\tag{e}$$
Where the third and fourth terms are just the first and second terms repeated, respectively, but with the positions of the $d$ and $c$ indices interchanged, as marked by the blue and red colors, respectively. But this is not what I see when I look at the full expression for $\nabla_c\nabla_dv^a$:
$$\nabla_c\nabla_dv^a=\underbrace{\partial_c\nabla_dv^a}_{(\mathrm{f})}-\underbrace{\Gamma_{dc}^b\nabla_b v^a}_{(\mathrm{g})}+\underbrace{\Gamma_{bc}^a\nabla_d v^b}_{(\mathrm{h})}$$
$$=\partial_c\left(\partial_dv^a+\Gamma_{bd}^av^b\right)-\Gamma_{dc}^b\left(\partial_bv^a+\Gamma_{eb}^av^e\right)+\Gamma_{bc}^a\left(\partial_dv^b+\Gamma_{ed}^bv^e\right)$$
$$=\partial_c\partial_dv^a+\left(\partial_c\Gamma_{bd}^a\right)v^b+\Gamma_{bd}^a\partial_cv^b\tag{f}$$
$$-\Gamma_{dc}^b\partial_bv^a-\Gamma_{dc}^b\Gamma_{eb}^av^e\tag{g}$$
$$+\Gamma_{bc}^a\partial_dv^b+\Gamma_{bc}^a\Gamma_{ed}^bv^e\tag{h}$$
$$=\fbox{$\partial_c\partial_dv^a+\left(\partial_c\Gamma_{bd}^a\right)v^b-\Gamma_{dc}^b\partial_bv^a-\Gamma_{dc}^b\Gamma_{eb}^av^e+\Gamma_{bc}^a\Gamma_{ed}^bv^e+\Gamma_{bd}^a\partial_cv^b+\Gamma_{bc}^a\partial_dv^b$}$$
The labelled eqns ($\mathrm{f}$-$\mathrm{h}$) show the expansion of each of the terms above in the two equations above. There are in fact $7$ terms, the last two terms are symmetric in the $c$ and $d$ indices, that is, swapping the $c$ and $d$ indices in the 6th term of the boxed expression gives the 7th and final term, namely, $\Gamma_{bc}^a\partial_d 
v^b$.
By comparing the boxed equation for $\nabla_c\nabla_dv^a$ and eqn $(\mathrm{c})$, in fact, the only two terms common to both the boxed eqn and $(\mathrm{c})$ are the 2nd and 5th terms in the boxed eqn. So what happened to the other terms? More importantly, why did the 1st term, $\partial_c\partial_dv^a$ disappear? I can't invoke $\partial_c\partial_dv^a-\partial_d\partial_cv^a=0$ as explained in the answer as I have not subtracted off $\nabla_d\nabla_cv^a$ yet.
For the sake of completeness I will now compute $\nabla_d\nabla_cv^a$ (though this is only for reference and is not needed to answer the questions I have here):
$$\nabla_d\nabla_cv^a=\underbrace{\partial_d\nabla_cv^a}_{(\mathrm{i})}-\underbrace{\Gamma_{cd}^b\nabla_b v^a}_{(\mathrm{j})}+\underbrace{\Gamma_{bd}^a\nabla_c v^b}_{(\mathrm{k})}$$
$$=\partial_d\left(\partial_cv^a+\Gamma_{bc}^av^b\right)-\Gamma_{cd}^b\left(\partial_bv^a+\Gamma_{eb}^av^e\right)+\Gamma_{bd}^a\left(\partial_cv^b+\Gamma_{ec}^bv^e\right)$$
$$=\partial_d\partial_cv^a+\left(\partial_d\Gamma_{bc}^a\right)v^b+\Gamma_{bc}^a\partial_dv^b\tag{i}$$
$$-\Gamma_{cd}^b\partial_bv^a-\Gamma_{cd}^b\Gamma_{eb}^av^e\tag{j}$$
$$+\Gamma_{bd}^a\partial_cv^b+\Gamma_{bd}^a\Gamma_{ec}^bv^e\tag{k}$$
$$=\fbox{$\partial_d\partial_cv^a+\left(\partial_d\Gamma_{bc}^a\right)v^b-\Gamma_{cd}^b\partial_bv^a-\Gamma_{cd}^b\Gamma_{eb}^av^e+\Gamma_{bd}^a\Gamma_{ec}^bv^e+\Gamma_{bc}^a\partial_dv^b+\Gamma_{bd}^a\partial_cv^b$}$$
The eqns ($\mathrm{i}$-$\mathrm{k}$) simply represent expanded forms of the each of the terms in the equations above.
I didn't want to resort to doing this and it was tedious to type this all out in full, but I did it for a reason: I am beginning to doubt myself on the validity of the boxed expressions I have worked out. For instance, what is to stop me writing the last term of the 2nd boxed expression as $\Gamma_{cd}^b\partial_bv^a$? This term has the same contravariant index as the LHS, $\nabla_d\nabla_cv^a$, namely, $a$ and the same covariant indices, $d$ and $c$ on the LHS.
So I know this may seem like a question in its' own right, but I consider it to be pertinent to this post, so does this mean that $\Gamma_{bd}^a\partial_cv^b=\Gamma_{cd}^b\partial_bv^a$? I never thought they could be equal but now I'm not so sure as the live indices are all preseved.

Final remarks:
Due to work commitments, I was unable to respond to this post or even self-study until now. My apologies for this and for an excessively long post, I laboured the point somewhat to convey my confusion in the most effective way I could think of. For these reasons I think that placing a bounty on this question is the least I could do.

2nd update addressing recent answers/comments:
I see some very nice answers emerging, which, for the most part answers my main question; how to obtain eqn $(\mathrm{c})$ from $(\mathrm{b})$. I would like to thank all participants for your kind help so far.
One of the questions I asked that did go overlooked, was

What is to stop me from writing the last term of the 2nd boxed expression as $\Gamma_{cd}^b\partial_bv^a$? This term has the same contravariant index as the LHS, $\nabla_d\nabla_cv^a$, namely, $a$ and the same covariant indices, $d$ and $c$ on the LHS?

In short, does the 7th (final) term of the second boxed expression satisfy $$\Gamma_{bd}^a\partial_cv^b=\Gamma_{cd}^b\partial_bv^a\tag{?}$$
If that was too vague, then put another way, am I at liberty to switch around whichever tensor indices I like (even if they belong to differential operators, etc), with the caveat that I preserve the overall labelling of the contravariant and covariant indices? If any term in the expression on the LHS or RHS has a contravariant index $a$ and covariant indices $d$ and $c$ as long as this is maintained throughout all other terms may I switch the indices to different factors of the terms ($\Gamma_{bd}^a\partial_cv^b=\Gamma_{cd}^b\partial_bv^a$)?
 A: When there is no torsion the "gammas" are symmetric in the covariant indices
$$\Gamma_{h\,\,\,k}^{\,\,\,j}=\Gamma_{k\,\,\,h}^{\,\,\,j}$$
That means that the term
$$\Gamma_{d\,\,\,c}^{\,\,\,b}(\partial_bv^a+\Gamma_{e\,\,\,b}^{\,\,\,a}v^e)$$
is symmetric in $c$ and $d$.
Symmetric terms in $c$ and $d$ will cancel out when we subtract $\nabla_d\nabla_cv^a$ from $\nabla_c\nabla_dv^a$.
The ordinary partial derivatives
$$\partial_d\partial_cv^a=\partial_c\partial_dv^a$$
are also symmetric in $c$ and $d$ since $v^a(x^j)\in C^2$ (see Wikipedia on Symmetry of second derivatives)
Finally, you need to use the product rule (Leibniz rule)
$$\partial_c(\Gamma_{b\,\,\,d}^{\,\,\,a}v^b)=\partial_c(\Gamma_{b\,\,\,d}^{\,\,\,a})v^b+\Gamma_{b\,\,\,d}^{\,\,\,a}\partial_cv^b$$
which means you will have an extra term compared to your calculations
$$\nabla_c\nabla_dv^a=\partial_c\partial_dv^a+\partial_c(\Gamma_{b\,\,\,d}^{\,\,\,a})v^b+\Gamma_{b\,\,\,d}^{\,\,\,a}\partial_cv^b-\Gamma_{d\,\,\,c}^{\,\,\,b}(\dots)+\Gamma_{b\,\,\,c}^{\,\,\,a}\partial_dv^b+\Gamma_{b\,\,\,c}^{\,\,\,a}\Gamma_{e\,\,\,d}^{\,\,\,b}v^e$$
Now we see that the expression $(\Gamma_{b\,\,\,d}^{\,\,\,a}\partial_cv^b+\Gamma_{b\,\,\,c}^{\,\,\,a}\partial_dv^b)$ is obviously symmetric in $c$ and $d$.
So in conclusion
$$\nabla_c\nabla_dv^a=\partial_c(\Gamma_{b\,\,\,d}^{\,\,\,a})v^b+\Gamma_{b\,\,\,c}^{\,\,\,a}\Gamma_{e\,\,\,d}^{\,\,\,b}v^e+\textit{symmetric contributions}$$
Update
The author is collecting symmetric terms under the umbrella "symmetric contributions" to make the practical computations easier. It's a way of being "lazy". But how does it work? Lets  find out! Our goal is to figure out
\begin{equation}
(\nabla_c\nabla_d-\nabla_d\nabla_c)v^a {\tag 1}
\end{equation}
This is actually a measure of the symmetry properties- or rather, the lack thereof- of repeated partial covariant differentiation.
An entity is said to be symmetric in a pair of superscripts (or in a pair of subscripts) if an interchange of the indices concerned does not affect the values of the components.
So for instance, the mixed partial derivatives
\begin{equation}
\partial _c\partial _d v^a=\frac{\partial^2 v^a}{\partial x^c \partial x^d}=\frac{\partial^2 v^a}{\partial x^d \partial x^c}=\partial_d\partial_c v^a{\tag 2}
\end{equation}
is an example of an entity where an interchange of the indices $(c,d)$ does not change the values of the components.
Not all of the terms in $\nabla_c\nabla_d v^a$ are symmetric, but those who are will cancel out when we put everything together in the expression $(\nabla_c\nabla_d-\nabla_d\nabla_c)v^a$
In your updated question you have correctly and laboriously calculated two boxed expressions, one for $\nabla_c \nabla_dv^a$ and one for $\nabla_d\nabla_cv^a$. These represent the complete calculation which means you have already done all the work and there is no need to use this "symmetric contribution"-shenanigans to simplify the calculations. But we can use your calculations to illustrate how symmetric terms cancel each other in a pairwise fashion. I will now go through the symmetric terms, one by one.
The first term in your first box is $\partial_c\partial_d v^a$, the first term in your second box is $\partial_d\partial_c v^a$. Upon forming the expression $(\nabla_c\nabla_d-\nabla_d\nabla_c)v^a$ we see that this part gives us the contribution
\begin{equation}
\partial_c \partial_dv^a -\partial_d\partial_cv^a=\partial_c \partial_dv^a -\partial_c\partial_dv^a=0
\tag{3}
\end{equation}
Notice how the "total" symmetric contribution is zero. The third and fourth terms in your first box are $-\Gamma^b_{dc}(\partial_b v^a+\Gamma^a_{eb}v^e)$. The third and fourth terms in your second box are $-\Gamma^b_{cd}(\partial_b v^a+\Gamma^a_{eb}v^e)$. Again, when we form the expression $(\nabla_c\nabla_d-\nabla_d\nabla_c)v^a$ we see that this part gives us the contribution (remember that $\Gamma^b_{cd}=\Gamma^b_{dc}$)
\begin{equation}
-\Gamma^b_{dc}(\partial_b v^a+\Gamma^a_{eb}v^e)-(-\Gamma^b_{cd}(\partial_b v^a+\Gamma^a_{eb}v^e))=0
\tag{4}\end{equation}
Finally, the sixth and seventh terms in your first box are $\Gamma^a_{bd}\partial_cv^b+\Gamma^a_{bc}\partial_dv^b$. The sixth and seventh terms in your second box are $\Gamma^a_{bc}\partial_dv^b+\Gamma^a_{bd}\partial_cv^b$. This gives us the contribution
\begin{equation}
(\Gamma^a_{bd}\partial_cv^b+\Gamma^a_{bc}\partial_dv^b)-(\Gamma^a_{bc}\partial_dv^b+\Gamma^a_{bd}\partial_cv^b)=0\tag{5}
\end{equation}
In conclusion $\partial_c \partial_dv^a$, $-\Gamma^b_{dc}(\partial_b v^a+\Gamma^a_{eb}v^e)$ and  $(\Gamma^a_{bd}\partial_cv^b+\Gamma^a_{bc}\partial_dv^b)$ are symmetric in the indices $(c,d)$ and hence cancel out. This is why the mystical "+symmetric contributions" disappear in the final result.
We are left with the second and fifth term in your first box minus the second an the fifth term in your second box
\begin{equation}
(\nabla_c\nabla_d-\nabla_d\nabla_c)v^a=\partial_c\Gamma^a_{bd}v^b+\Gamma^a_{bc}\Gamma^b_{ed}v^e-\partial_d\Gamma^a_{bc}v^b- \Gamma^a_{bd}\Gamma^b_{ec}v^e\tag{6}
\end{equation}
Renaming dummy index $b\leftrightarrow e$ and putting it in the usual order
\begin{equation}
(\nabla_c\nabla_d-\nabla_d\nabla_c)v^a=\left(\partial_c\Gamma^a_{bd}-\partial_d\Gamma^a_{bc}+\Gamma^a_{ec}\Gamma^e_{bd}- \Gamma^a_{ed}\Gamma^e_{bc}\right)v^b\tag{7}
\end{equation}
A: First, since you've already gone most of the way to the fully written-out proof, let's just finish it. Then we can look back in hindsight at what the author was doing by not fully writing it out.
You correctly showed (first boxed equation) that $\nabla_c\nabla_d v^a$ equals
$$
\partial_c\partial_dv^a+\left(\partial_c\Gamma_{bd}^a\right)v^b-\Gamma_{dc}^b\partial_bv^a-\Gamma_{dc}^b\Gamma_{eb}^av^e+\Gamma_{bc}^a\Gamma_{ed}^bv^e+\Gamma_{bd}^a\partial_cv^b+\Gamma_{bc}^a\partial_dv^b.
$$
You also correctly showed (second boxed equation) that $\nabla_d\nabla_c v^a$ equals
$$
\partial_d\partial_cv^a+\left(\partial_d\Gamma_{bc}^a\right)v^b-\Gamma_{cd}^b\partial_bv^a-\Gamma_{cd}^b\Gamma_{eb}^av^e+\Gamma_{bd}^a\Gamma_{ec}^bv^e+\Gamma_{bc}^a\partial_dv^b+\Gamma_{bd}^a\partial_cv^b.
$$
Since $\partial_d\partial_cv^a = \partial_c\partial_dv^a$ and $\Gamma_{cd}^b = \Gamma_{db}^b$, most terms cancel out when we compute the difference $\nabla_c\nabla_d v^a - \nabla_d\nabla_c v^a$. The difference is
$$
(\partial_c\Gamma_{bd}^a)v^b
- (\partial_d \Gamma_{bc}^a)v^b
+ \Gamma_{bc}^a\Gamma_{ed}^b v^e
- \Gamma_{bd}^a\Gamma_{ec}^b v^e.
$$
After relabeling the dummy indices in the last two terms, we obtain $(\partial_c\Gamma_{bd}^a - \partial_d \Gamma_{bc}^a + \Gamma_{ec}^a\Gamma_{bd}^e - \Gamma_{ed}^a\Gamma_{bc}^e) v^b$ which gives the desired expression for $R_{bcd}^a$.

If you followed the full proof all the way through, now the author's more abbreviated proof may make more sense in hindsight. Put yourself in the mindset of

*

*being a lazy person and not wanting to write down tedious details, and

*knowing ahead of time that the ultimate goal is not to compute $\nabla_c\nabla_d v^a$, but to compute $\nabla_c\nabla_d v^a - \nabla_d\nabla_c v^a$.

You compute $\nabla_c\nabla_d v^a$ which expands into seven terms, but you realize that five of them are going to cancel out later when you compute $\nabla_c\nabla_d v^a - \nabla_d\nabla_c v^a$. So you just write down the two terms that will matter and don't bother writing down the five terms that won't:
$$
\nabla_c\nabla_d v^a
= \left(\partial_c\Gamma_{bd}^a\right)v^b+\Gamma_{bc}^a\Gamma_{ed}^bv^e + \text{stuff which we're too lazy to write down}
$$
We are not saying that the five terms we didn't write down are actually equal to $0$. They aren't. We're saying that they will cancel out later when we compute $\nabla_c\nabla_d v^a - \nabla_d\nabla_c v^a$. Being "symmetric in $c$ and $d$" is a terse way to explain why those terms will cancel out later.
A: In the original posting you have already calculated (cf (f),(g),(h),
the term in the box) :
$$\nabla_c\nabla_d v^a =
\partial_c\partial_d v^a +
\partial_c\Gamma_{bd}^a\ v^b -
\Gamma_{dc}^b\partial_b v^a - \Gamma_{dc}^b \Gamma_{eb}^a v^e +
\Gamma_{bc}^a \Gamma_{ed}^b v^e + \Gamma_{bd}^a
\partial_cv^b +\Gamma_{bc}^a \partial_d v^b $$
(1) Here by the symmetry of index $c$ and $d$, the first, third, fourth
terms in LHS in $ \nabla_c\nabla_d v^a$ also exist in $ \nabla_d\nabla_c v^a$ so that they are
canceled.
(2) Consider the sum of the sixth and seventh terms i.e.
$$ \Gamma_{bd}^a
\partial_cv^b +\Gamma_{bc}^a \partial_d v^b$$ By symmetry, then the sum term also
exists in $ \nabla_d\nabla_c v^a$.
(3) Hence the only second and fifth terms in the LHS survive in $\nabla_c\nabla_d
v^a -\nabla_d\nabla_c v^a$, which finishes the proof.
[Add]
We have $$\nabla_c\nabla_d v^a =
 dx^a ( \nabla_c\nabla_d v  )
 = dx^a( \nabla_c( \nabla_d v) -\Gamma_{cd}^x
  \nabla_xv ) $$ by definition of $\nabla\nabla v$. By symmetry of index $c$ and $d$, we have
\begin{align*} \nabla_c\nabla_d v^a -\nabla_d\nabla_c v^a
 &=dx^a (\nabla_c( \nabla_dv) -
  \nabla_d( \nabla_c v  ) )
  \end{align*}

If we define a tensor
$T(\partial_c,\partial_d,dx^a)=\nabla_c\nabla_d v^a
-\nabla_d\nabla_c v^a $, then since $T$ is a tensor we don't have to
consider derivatives of $v^x$ because $T(\partial_c,\partial_d,dx^a)
$ does not contains derivatives of $v^x$. Now we must find a
multiple of a component $v^a$ in $\nabla_c( \nabla_dv)$ :

\begin{align*}dx^a( \nabla_c( \nabla_dv) ) &=
\partial_c( \nabla_dv^a ) + \Gamma_{cx}^a \nabla_d v^x
\\&= \partial_c (  \Gamma_{dx}^av^x ) +
\Gamma_{cx}^a \Gamma_{dy}^xv^y +S(c,d)
\\&=\partial_c \Gamma_{dx}^av^x  +
\Gamma_{cx}^a \Gamma_{dy}^xv^y +S'(c,d)
\end{align*} where each terms in $S,\ S'$ contains derivatives of $v^x$
