= Solution
Although $\nabla_\beta f=\partial_\beta f$ for a <scalar field>, its first derivative is a <covector field>. Therefore
$$
\nabla_\alpha\nabla_\beta f=\partial_\alpha\partial_\beta f-\Gamma^\gamma{}_{\alpha\beta}\partial_\gamma f.
$$
Commuting <partial derivatives> and using the <torsion-free connection> gives $\boxed{[\nabla_\alpha,\nabla_\beta]f=0}$. For a <vector field>, differentiate the $(1,1)$ <tensor> $\nabla_\beta V^\mu$ with both its indices included:
$$
\nabla_\alpha\nabla_\beta V^\mu
=\partial_\alpha(\nabla_\beta V^\mu)+\Gamma^\mu{}_{\alpha\epsilon}\nabla_\beta V^\epsilon
-\Gamma^\epsilon{}_{\alpha\beta}\nabla_\epsilon V^\mu.
$$
Subtract the expression with $\alpha,\beta$ exchanged. Second derivatives of $V$ cancel; the terms multiplying first derivatives cancel pairwise, and the final lower-index connection terms cancel by symmetry. The remaining coefficient of $V^\nu$ is
$$
\boxed{[\nabla_\alpha,\nabla_\beta]V^\mu=
\left(\partial_\alpha\Gamma^\mu{}_{\beta\nu}-\partial_\beta\Gamma^\mu{}_{\alpha\nu}
+\Gamma^\mu{}_{\alpha\epsilon}\Gamma^\epsilon{}_{\beta\nu}-\Gamma^\mu{}_{\beta\epsilon}\Gamma^\epsilon{}_{\alpha\nu}\right)V^\nu
=R^\mu{}_{\nu\alpha\beta}V^\nu.}
$$
This is the <curvature commutator>. On a covariant index the sign reverses, giving the <curvature commutator on a covariant tensor> used in the next part.
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