= Solution
Use sequential semicolon indices: $X^a{}_{;kl}=\nabla_l(\nabla_kX^a)$, where the second <covariant derivative> also differentiates the lower index of $\nabla_kX^a$. For a <torsion-free connection>, direct expansion gives
$$
X^a{}_{;kl}=X^a{}_{,kl}+\Gamma^a{}_{bk,l}X^b+\Gamma^a{}_{bk}X^b{}_{,l}
+\Gamma^a{}_{bl}X^b{}_{,k}+\Gamma^a{}_{pl}\Gamma^p{}_{bk}X^b
-\Gamma^p{}_{kl}X^a{}_{,p}-\Gamma^p{}_{kl}\Gamma^a{}_{bp}X^b.
$$
Subtract the expression with $k,l$ exchanged. Commuting partial derivatives cancel the first term; the first-derivative terms cancel in pairs; symmetry of the lower <Christoffel symbol> indices cancels both terms containing $\Gamma^p{}_{kl}$. What remains is
$$
X^a{}_{;kl}-X^a{}_{;lk}
=\left(\Gamma^a{}_{bk,l}-\Gamma^a{}_{bl,k}
+\Gamma^a{}_{pl}\Gamma^p{}_{bk}-\Gamma^a{}_{pk}\Gamma^p{}_{bl}\right)X^b
=R^a{}_{bkl}X^b.
$$
This is the <Ricci identity> with the curvature convention printed on the PDF's first page. In terms of derivative operators it is $[\nabla_l,\nabla_k]X^a=R^a{}_{bkl}X^b$; writing the operator order explicitly avoids reversing the sign.
The lower-index rule follows by applying the commutator to the scalar contraction $Y_aX^a$. Its second derivatives commute, so the <product rule> implies $Y_{a;kl}-Y_{a;lk}=-R^p{}_{akl}Y_p$. The derivative commutator is a derivation on tensor products, since the cross terms cancel. It therefore acts once on every index, with a positive sign on each upper index and a negative sign on each lower index. For an arbitrary <tensor> of valence $(r,s)$,
$$
\boxed{
T^{a_1\cdots a_r}{}_{b_1\cdots b_s;kl}
-T^{a_1\cdots a_r}{}_{b_1\cdots b_s;lk}
=\sum_{\alpha=1}^rR^{a_\alpha}{}_{pkl}
T^{a_1\cdots p\cdots a_r}{}_{b_1\cdots b_s}
-\sum_{\beta=1}^sR^p{}_{b_\beta kl}
T^{a_1\cdots a_r}{}_{b_1\cdots p\cdots b_s}.}
$$
Local sums of decomposable tensors establish the formula for every <tensor field>, rather than only for a single tensor product.
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