Does it make sense to ask how the covariant derivative act on the partial derivative $\nabla_\mu ( \partial_\sigma)$? If so, what is the answer? I want to find out how the covariant derivative acts on terms containing a partial derivative, e.g. $ \nabla_\mu(k^\sigma\partial_\sigma l_\nu)$. But I don't know how to evaluate the terms of the form $\nabla_\mu(\partial_\sigma)$. If one writes
$$
\nabla_\mu(k^\sigma\partial_\sigma l_\nu) = \nabla_\mu(g^{\rho \sigma}k_\rho\partial_\sigma l_\nu) = g^{\rho \sigma}\nabla_\mu(k_\rho\partial_\sigma l_\nu) = g^{\rho \sigma}\left[ \nabla_\mu(k_\rho)\partial_\sigma l_\nu + k_\rho\nabla_\mu(\partial_\sigma )l_\nu + k_\rho\partial_\sigma\nabla_\mu( l_\nu) \right]
$$
Problem: How to determine $\nabla_\mu(\partial_\sigma )$. How do I work it out, and understand whatever the answer, that it makes sense? Have I made a mistake?
EDIT:
I add the context: suppose $k^a$ and $l^a$ are killing vector. Then I want to prove that the commutator $[k,l]_\alpha = k^\sigma\partial_\sigma l_\alpha - l^\sigma\partial_\sigma k_\alpha$ is a Killing vector. If you write out $\nabla_{(\mu}[k,l]_{\nu)}$, then you find these terms immediately. 
 A: My question would be why are you doing this?  
The idea of the covariant derivative is that it maps a tensor to a tensor with one more lowered index, while satisfying a few other rules like the Liebniz rule.  
But, if you have an object like $\partial_{a}\ell_{b}$, it is already, in general, not a tensor, and your mapping has a domain problem.
If you're trying to figure out the expression of some series of covariant derivatives in terms of partials and Christoffel symbols, you need to do something like:
$$
\nabla_{a}\nabla_{b}v^{c} = \partial_{a}\left(\nabla_{b}v^{c}\right) - \Gamma_{ab}{}^{d}\nabla_{d}v^{c} + \Gamma_{ad}{}^{c}\nabla_{b}v^{d}\\
$$
And then you can expand out the $\nabla v$ terms normally.  But it doesn't mean anything to compute the covariant derivative of a partial, except in one case:  when you are computing the derivative of a function.  In that case, we have, for all $f$, $\nabla_{a}f = \partial_{a}f$ by definition, so it doesn't matter which one you use.
A: OP asks in the title (v2):

Does it make sense to ask how the covariant derivative acts on the partial derivative $\nabla_\mu ( \partial_\sigma)$?

I) Well, Yes, in a limited sense, if one is careful with the notation. Recall that a partial derivative $\partial_{\mu}$ may be interpreted as serving a double purpose in differential geometry: both as an actual derivative $\frac{\partial}{\partial x^{\mu}}$ acting on functions, or merely as a booking device, as a basis, that transforms correctly under change of local coordinates $x^{\mu}$. To make this distinction clear, let us introduce the notation $b_{\mu}$ for the latter role. Then we may e.g. write a vector field as 
$$X~=~X^{\mu}(x)b_{\mu}.\tag{1}$$ 
We can then reproduce covariant differentiation 
$$X^{\nu}_{;\mu}~=~\frac{\partial X^{\nu}}{\partial x^{\mu}} + \Gamma^{\nu}_{\mu\lambda} X^{\lambda}\tag{2}$$
via a trick: Introduce the formal first-order differential operator
$$ \nabla_{\mu}~=~\frac{\partial}{\partial x^{\mu}}  + \Gamma^{\nu}_{\mu\lambda}b_{\nu} \frac{\partial}{\partial b_{\lambda}}.\tag{3}$$
Then
$$\nabla_{\mu}X~\stackrel{(1)+(3)}{=}~\left(\frac{\partial}{\partial x^{\mu}}  + \Gamma^{\nu}_{\mu\lambda}b_{\nu} \frac{\partial}{\partial b_{\lambda}}\right)(X^{\kappa}b_{\kappa})~=~\left(\frac{\partial X^{\nu}}{\partial x^{\mu}} + \Gamma^{\nu}_{\mu\lambda} X^{\lambda}\right)b_{\nu}~\stackrel{(2)}{=}~X^{\nu}_{;\mu}b_{\nu}\tag{4}$$
II) Similarly, the Lie bracket $$[X,Y]^{\nu} ~=~ X^{\mu}\frac{\partial Y^{\nu}}{\partial x^{\mu}} -Y^{\mu}\frac{\partial X^{\nu}}{\partial x^{\mu}}\tag{5}$$ of vector fields 
$$X~=~X^{\mu}b_{\mu} , \qquad Y~=~Y^{\mu}b_{\mu}, \tag{6} $$
can be reproduced by the Schouten-Nijenhuis bracket
$$[X,Y]~=~X\left(\stackrel{\leftarrow}{\frac{\partial}{\partial b_{\mu}}}\stackrel{\rightarrow}{\frac{\partial}{\partial x^{\mu}}} -  \stackrel{\leftarrow}{\frac{\partial}{\partial x^{\mu}}}\stackrel{\rightarrow}{\frac{\partial}{\partial b_{\mu}}} \right)Y. \tag{7}$$
III) Such formalism, which differentiates basis elements $b_{\mu}$, can be developed further in other areas of differential geometry.
A: If you want to use this for commutators, then either consider
$$\nabla_\sigma([k,l]^\mu)=\partial_\sigma[k,l]^\mu+\Gamma^\mu_{\sigma\kappa}[k,l]^\kappa=\partial_\sigma(k^\nu\partial_\nu l^\mu-l^\nu\partial_\nu k^\mu)+\Gamma^\mu_{\sigma\kappa}(k^\nu\partial_\nu l^\kappa-l^\nu\partial_\nu k^\kappa)... $$
and use this for further calculations, or consider that for any torsionless connection, we have $$ [k,l]^\mu=k^\nu\partial_\nu l^\mu-l^\nu\partial_\nu k^\mu\equiv k^\nu\nabla_\nu l^\mu-l^\nu\nabla_\nu k^\mu, $$ and the latter expression contains only terms that are 'covariant'.
