Is it okay to relabel tensor indices as long as the overall contravariant and covariant indices on each term in a tensor expression are the same? I wanted to ask this in my previous question as I thought it was more appropriately placed given that all the context is present there, but a fellow user has suggested that it is better to ask it as a separate question. So that's what I do now.

Suppose we have a tensor expression like $$\nabla_d\nabla_cv^a=\partial_d\partial_cv^a+\left(\partial_d\Gamma_{bc}^a\right)v^b-\Gamma_{cd}^b\partial_bv^a-\Gamma_{cd}^b\Gamma_{eb}^av^e+\Gamma_{bd}^a\Gamma_{ec}^bv^e+\Gamma_{bc}^a\partial_dv^b+\color{red}{\Gamma_{bd}^a\partial_cv^b}$$
Looking at the final term (marked red) in the expression above, is there anything preventing me from re-writing that term as say, $\Gamma_{cd}^b\partial_bv^a$?
This is the same as asking; Is it enough to ensure that each term in an expression has the same contravariant and covariant indices as the LHS, namely, with contravariant index $a$ and covariant indices $c$ and $d$?
A third way of asking this question is
$$\Gamma_{bd}^a\partial_cv^b\stackrel{?}{=}\Gamma_{cd}^b\partial_bv^a\tag{b}$$

My first thought is that the answer is no; there is nothing to stop me from changing the positions of the indices whether they belong to operators, tensors, vectors, etc. All is okay as long as the same contravariant and covariant indices are present in each term on the RHS to match the LHS.
What confuses me however is in the context of the previous question I asked, I was required to subtract the expression, $(\mathrm{a})$ from a similar expression for $\nabla_c\nabla_dv^a$. But if my intuition is right and I am at liberty to swap indices, like in $(\mathrm{b})$, (as long as the overall contravariant and covariant indices are preserved), then I will not get these symmetric contributions to cancel as the indices will not match - please see this previous question for what I say here to make sense.
Can someone please explain what I am not understanding here?
 A: Why would you be allowed to arbitrarily change indices for a single term on the RHS? Equation (b) is a-priori not true (it’s true for specific choices of $\Gamma$, for example the flat connection in a specific coordinate system) but I don’t see why it holds in general.
Let’s take a specific example. Fix an arbitrary connection $\nabla$ on the tangent bundle, and a coordinate system. Then, the torsion of $\nabla$ is a $(1,2)$-tensor field which has components
\begin{align}
T^a_{\,bc}&=\Gamma^a_{\,bc}-\Gamma^a_{\,cb}.\tag{$*$}
\end{align}
Now, you can’t just arbitrarily say “the LHS has index $a$ upstairs and indices $b,c$ downstairs so we also have $T^a_{\,bc}=\Gamma^a_{cb}-\Gamma^a_{bc}$”. In this second equation, both sides still have index $a$ upstairs and indices $b,c$ downstairs; but this is total nonsense (if this equation were true, we’d find that the torsion is zero, which is not true in general).
What you are allowed to do is change all the indices everywhere:
\begin{align}
\begin{cases}
T^{a}_{\,cb}&=\Gamma^a_{\,cb}-\Gamma^a_{\,bc}\\
T^{@}_{\,\sharp\flat}&=\Gamma^@_{\,\sharp\flat}-\Gamma^@_{\,\flat\sharp}\\
T^{\alpha}_{\,\beta\delta}&=\Gamma^{\alpha}_{\,\beta\delta}-\Gamma^{\alpha}_{\,\delta\beta},
\end{cases}
\end{align}
where $a,b,c,@,\sharp,\flat,\alpha,\beta,\delta\in\{1,\dots, n\}$. These are all examples of valid equations which follow directly from the definition $(*)$. Here, I have made the appropriate changes everywhere, not just shuffling around indices for a few specific terms.

Edit:
I just took a look at your link and you already have several great answers, so I’m not really sure what the issue is. Anyway, here’s what I’ll say. Let us give definitions for certain sub-terms:
\begin{align}
\nabla_d\nabla_cv^a=\underbrace{\partial_d\partial_cv^a}_{A^a_{cd}}+\underbrace{\left(\partial_d\Gamma_{bc}^a\right)v^b}_{B^a_{cd}}+\underbrace{(-\Gamma_{cd}^b\partial_bv^a-\Gamma_{cd}^b\Gamma_{eb}^av^e)}_{C^{a}_{cd}}+
\underbrace{\Gamma_{bd}^a\Gamma_{ec}^bv^e}_{=D^a_{cd}}+
\underbrace{(\Gamma_{bc}^a\partial_dv^b+\Gamma_{bd}^a\partial_cv^b)}_{E^a_{cd}}.
\end{align}
So, $\nabla_d\nabla_cv^a=A^a_{cd}+B^{a}_{cd}+C^a_{cd}+D^a_{cd}+E^a_{cd}$ is a sum of five terms. Now, by inspecting the definitions, the following is very obvious: $A,C,E$ are symmetric in the lower two indices. We have $A^a_{cd}=A^a_{dc}$ due to equality of mixed partials. We have $C^a_{cd}=C^a_{dc}$ because you’re assuming a torsion-free connection in that post. Lastly, $E^a_{cd}=E^a_{dc}$ is very clear (if you want, the reason is commutatitivty of addition of real numbers).
So, out of thse five terms, three of them are symmetric in the lower indices $c,d$, so if you subtract $\nabla_{c}\nabla_dv^a$, then only the $B,D$ terms will contribute:
\begin{align}
\nabla_d\nabla_cv^a-\nabla_c\nabla_dv^a&=(B^a_{cd}-B^a_{dc})+(D^a_{cd}-D^a_{dc})\\
&=(\partial_d\Gamma^a_{bc}v^b-\partial_c\Gamma^a_{bd}v^b)+ (\Gamma_{bd}^a\Gamma_{ec}^bv^e-\Gamma_{bc}^a\Gamma_{e}^bv^e).
\end{align}
Now, you can rename some dummy indices to figure out what the components of the Riemann curvature are.
