Why is the partial derivative operator a map from $(k,l)$ tensor fields to $(k, l+1)$ tensor fields? On Section 3.2 in Carroll's Spacetime and Geometry, he claims the following:

In flat space in inertial coordinates, the partial derivative operator $\partial_\mu$ is a map from $(k,l)$ tensor fields to $(k, l+1)$ tensor fields, which acts linearly on its arguments and obeys the Leibniz rule on tensor products.

I have been wondering why is that so for quite some time and could not get a clue (it is the first time I am studying differential geometry).
My first attempt was to take a simple and practical example and understand it working it out. So I took a $(0,0)$ tensor (a scalar $\phi$), and applied the partial derivative operator to it, yielding $\partial_\mu \phi = \dfrac{\partial\phi}{\partial x^\mu}$. By hypothesis, the result supposedly should be a $(0,1)$ tensor, but by definition a $(0,1)$ tensor should look like $\omega = \omega_\mu dx^\mu$.
I have a feeling that I am really close to understanding it, but I was not able to get closure so far. Could any of you help me finding what am I missing on this specific example, or perhaps to give reasoning to the more general statement regarding partial derivative operators on tensor fields?
 A: In the Physics literature, it is quite common to interchange the terms "tensor" and "components of a tensor". That is pretty much the issue in here. While Carroll's statement is correct, the notation (which is quite common in Physics) can make the interpretation a bit more complicated.
Let us recall that a $(0,1)$ tensor field is, also by definition, a map from the space of $(1,0)$ tensor fields to the space of scalar functions. In other words, it "eats" vector fields and "spits" functions. $\partial_\mu \phi$ does this, since it can be fed a vector $X^\mu$ to yield the directional derivative of $\phi$ in the direction of $X^\mu$: $X^\mu \partial_\mu \phi$. The notation $\partial_\mu$ makes this a bit difficult to notice if you are thinking of tensors instead of tensor components (notice that, in components, it makes sense), so some other authors often choose to write $\partial_X$, where the vector field is already fed into the partial derivative (this is most often seen as $\nabla_X$, when using covariant derivatives). In this case, $\partial_X \phi$ is a function.
It holds that $\partial_{X + \alpha Y} \phi = \partial_X + \alpha \partial_Y \phi$, so it often ends up being convenient for calculations to write $X^\mu\partial_\mu$ instead of $\partial_X$.
In short, its pretty much a matter of notation. Thinking of tensors in terms of their components makes this statement in this particular notation a bit easier to understand. Essentially, $\partial_\mu$ takes a tensor with components $T^{\nu_1 \ldots \nu_k}{}_{\sigma_1 \ldots \sigma_l}$ in some coordinate system to the tensor with components $\partial_\mu T^{\nu_1 \ldots \nu_k}{}_{\sigma_1 \ldots \sigma_l}$ in the same coordinate system.
