Abstract formulation of the Riemann-tensor in index notation Let $M$ be a smooth manifold and let $g$ be a Riemannian or pseudo-Riemannian metric on it, with $\nabla$ being the associated Levi-Civita connection.
The indices $a,b,c,d...$ denote abstract indices (but component indices would suffice as well).
While the usual expression for the Riemann-tensor involves a long expression containing partial derivatives and connection coefficients, it is possible to give an abstract formulation more similar to the indexless expression for the Riemann-tensor as $$ R^a_{\ bcd}X^b=[\nabla_c,\nabla_d]X^a. $$
This is obviously an operational instruction on how the Riemann-tensor acts on $X$, and due to the unusual way the covariant derivative interacts with indices, the vector field cannot be decoupled from this expression. However, $X^a=\delta^a_{\ b}X^b$, and the covariant derivative of $\delta$ is zero, thus $$ R^a_{\ bcd}X^b=[\nabla_c,\nabla_d](\delta^a_{\ b}X^b)=\delta^a_{\ b}[\nabla_c,\nabla_d]X^b, $$ and from this we can decouple $X$, and we get $$ R^a_{\ bcd}=\delta^a_{\ b}[\nabla_c,\nabla_d]. $$
I have never seen this expression before like this, and also I can see a fundamental issue with this. All information regarding the Riemann-tensor is already contained in $[\nabla_c,\nabla_d]$, and the Kronecker-delta is only required to have the proper index structure. This expression is also manifestly skew-symmetric in $c$ and $d$, but the Riemann-tensor is also skew in $a$ and $b$, and the expression $$ \delta^a_{\ b}[\nabla_c,\nabla_d] $$ is manifestly symmetric in $a$ and $b$.
I realize that this might be due to the fact that while this expression looks like the tensor product of the identity tensor with the commutator of covariant derivatives, but since the commutator is not a tensor field, this is not true, and when we contract this expression with $b$ we not only interact with the delta tensor, but also with the derivative operators, however I don't see any invalid moves I would have performed in deriving this, yet this form is symmetric in the first two indices, but the Riemann-tensor is skew in the first two indices.
Question: Where is the problem with this line of thought?
 A: The equation
$$ R^a{}_{bcd}X^b = \delta^a_b[\nabla_c,\nabla_d]X^b $$
is correct - the issue is when you "equate coefficients of $X^b$".
When you have  $T_b X^b = S_b X^b$ for some tensors $T,S$ and all $X$, equating coefficients is legitimate:  set $X = \partial_i$ and you have $(T_b - S_b) \delta_i^b = 0$; i.e. $T_i = S_i$ for each $i$. 
In this case, however, what we have is not just a product of tensors, and the derivative operators mean we can't divide by $X$. If we attempt to make the same kind of argument, we get to
$$ R^a{}_{icd}=\delta_b^a [\nabla_c, \nabla_d] \partial_i^b.$$
The LHS simplified as we want, but the RHS will not. Index notation is somewhat clumsy here - we would normally write $(\partial_i)^b = \delta_i^b$, but then it would look like we're taking the $(b,i)^\rm{th}$ component of a derivative of the (parallel!) Kronecker delta, when in fact we are taking the $b^\rm{th}$ component of a derivative of the vector $\partial_i$. Thus we cannot say
$$ \delta_b^a [\nabla_c, \nabla_d] \partial_i^b = \delta_i^a [\nabla_c, \nabla_d]. $$
Note that it is actually true that all information about the Riemann tensor is contained in $[\nabla_c, \nabla_d]$ - you just need to remember that even after fixing $c$ and $d$, this operator sends vector fields to vector fields. If you want to "factor out" the $X$ then you could write $$R^a{}_{bcd} = [\nabla_c,\nabla_d]_b^a,$$ but this is certainly not common notation - the definition is typically given as in your question or more abstractly as $R(X,Y) = [\nabla_X, \nabla_Y] - \nabla_{[X,Y]}$, and from there on we just use $R$.
