Covariant derivatives' commutator defines the Riemann tensor, viz. $[\nabla_\mu,\,\nabla_\nu]=R_{\mu\nu\rho\sigma}\nabla^\sigma$. Contracting the left-hand indices obtains $0$, as the Riemann tensor is antisymmetric in its leftmost indices. However, in general raising one index without contraction gives a nonzero commutator for the covariant derivatives. Of course, since covariant derivatives are tensors, so are their commutators.

The problem you've found with partial derivatives with upstairs indices actually provide a motivation for covariant derivatives, but we can explain that with the even simpler question of how these derivatives are defined. Does $\partial^\nu$, whatever that is, act by applying $\partial_\mu$ before or after multiplying by $g^{\mu\nu}$? Clearly, the problem is that partial derivatives are not "metric compatible", which In view of the Leibniz law means they don't annihilate the metric tensor, whereas covariant derivatives are designed to do exactly that.

Covariant derivatives' commutator defines the Riemann tensor, viz. $[\nabla_\mu,\,\nabla_\nu]=R_{\mu\nu\rho\sigma}\nabla^\sigma$. Contracting the left-hand indices obtains $0$, as the Riemann tensor is antisymmetric in its leftmost indices. However, in general raising one index without contraction gives a nonzero commutator for the covariant derivatives. Of course, since covariant derivatives are tensors, so are their commutators.

The problem you've found with partial derivatives with upstairs indices actually provide a motivation for covariant derivatives, but we can explain that with the even simpler question of how these derivatives are defined. Does $\partial^\nu$, whatever that is, act by applying $\partial_\mu$ before or after multiplying by $g^{\mu\nu}$? Clearly, the problem is that partial derivatives are not "metric compatible", which in view of the Leibniz law means they don't annihilate the metric tensor, whereas covariant derivatives are designed to do exactly that.