Proving the Ricci identity

This seems like mainly a question about what the abstract index notation is asking you to do. Let's look at a single term from the right hand: $\nabla_X\nabla_YZ$. Recall that in abstract index notation, two tensors juxtaposed signifies their tensor product. For example, if $V^a$ and $W^b$ are vector fields, $V^aW^b$ is supposed to mean the tensor field $V \otimes W$. Pairing of an upper and lower index signifies a trace: if $\varphi_a$ is a 1-form and $V^b$ is a vector field, $\varphi_aV^a =\operatorname{tr}(\varphi\otimes V) = \varphi(V)$. So the way we're supposed to think of this is first taking their tensor product and then tracing. Lastly, notice that $\nabla_VW = \operatorname{tr}(\nabla V \otimes W)$ when considered as a section of $T^* \otimes T \otimes T$, where we trace over the first and last slots.

Returning to the first term above, we then have:

\begin{align} \nabla_X\nabla_YZ & = X^c\nabla_cY^d\nabla_dZ^b \\ & = \nabla_X(\operatorname{tr}(Y \otimes \nabla_-Z)) \\ & = \operatorname{tr}(\nabla_XY \otimes \nabla_-Z + Y \otimes \nabla_X\nabla_-Z)) \\ & = Y^dX^c \nabla_c\nabla_dZ^b + (X^c\nabla_cY)^d\nabla_dZ^b \\ & = \nabla^2_{X,Y}Z + \nabla_{\nabla_XY}Z. \end{align}

And from here the rest shouldn't be bad.

The start of the computation is OK, but the partial derivative does not make sense. I think the problems come from the fact that you misinterpret $\nabla_c\nabla_dZ^a$. Evaluating the second covariant derivative $\nabla^2Z$ on $X$ and $Y$, you do not get $\nabla_X\nabla_Y Z$. What you have to do is to differentiate $\nabla Z$ as a $\binom11$-tensor field. This implies that $(\nabla^2Z)(X,Y)=\nabla_X\nabla_YZ-\nabla_{\nabla_XY}Z$. Once you have this, torsion-freeness (in the standard form $\nabla_XY-\nabla_YX=[X,Y]$) easily implies the result.