Here is an attempt at an answer. Since we don't work with explicit values, but only in tensorial notation, we don't even care about the metric signature for our calculation. It is important only if you want to plug in some values.
Even in this case, see this question, where the accepted answer shows that the signature does not affect the Riemann tensor, or the Ricci tensor.
However, we must be careful about the definition of the Riemann Tensor. Almost everywhere I can remember it was defined as follows :
$$ [\nabla_\mu,\nabla_\nu]A^\rho = R^\rho_{\lambda\mu\nu}A^\lambda$$
However, some author might choose to define it with some indices distributed differently, which could make sign differences (for example if they define it with $\mu\leftrightarrow \nu$). Another thing to check, is the definition of the Ricci tensor. I usually use $R^\lambda_{\mu\lambda\nu} = R_{\mu\nu}$, but some authors might contract differently.
For your formula, with the definition of the Riemann tensor as I have given, here is the reasoning:
$$[\partial_\alpha,\Box]\phi = \partial_\alpha\Box\phi-\Box\partial_\alpha\phi$$
Now, there are two thing to notice. I can promote all derivative to covariant derivatives, since when they act on a scalar they are equivalent. Secondly, we have the following identity $\nabla_a\nabla_b\phi = \nabla_b\nabla_a\phi$. Try to prove it ! Using that :
$$\nabla_\alpha\Box\phi-\Box\nabla_\alpha\phi = \nabla_\alpha\nabla^\mu\nabla_\mu\phi-\nabla_\mu\nabla^\mu\nabla_\alpha\phi = \nabla_\alpha\nabla_\mu\nabla^\mu\phi-\nabla_\mu\nabla_\alpha\nabla^\mu\phi = [\nabla_\alpha,\nabla_\mu]\nabla^\mu\phi = R^\mu_{\sigma \alpha\mu}\nabla^\sigma\phi = -R^\mu_{\sigma\mu\alpha}\partial^\sigma\phi = -R_{\sigma\alpha}\partial^\sigma\phi$$
Since the expression is covariant, we can write it with upper or lower indices, it doesn't matter:
$$[\partial^\alpha,\Box]\phi = -R^{\alpha\sigma}\partial_\sigma\phi$$
