What do the $k$ and $k+1$ degrees in exterior derivatives count? Intuitively, differential $k$-forms in exterior derivatives could possibly (?) be approached as $(0,k)$ tensors with a constraint (antisymmetry).
A $1$-form, $\mathrm df$ will take a smooth function ($0$-form) and produce the directional derivative of $f$ at a point:
$$\underbrace{\quad\text{d}f\quad}_{\text{differential of a function}} =  \underbrace{\quad\frac{\partial f}{\partial x^i}\quad}_{\text{coefficients}}\quad\underbrace{\quad dx^i\quad}_{\text{basis of }V^*}$$
A $2$-form would take two vectors to produce a scalar:
$$\omega = \underbrace{\omega_{\mu\nu}}_{\text{components}}\;\underbrace{ dx^\mu \otimes dx^\nu }_{\text{basis}}$$
The question:

How can the ideas above be connected to the sentence in Wikipedia:
"The exterior derivative of a differential form of degree $k$ is a differential form of degree $k + 1.$"?
Or in other words, in the examples at the beginning of the post, what would be $k$ and what would be $k+1$?

 A: If $\omega$ is a $k$-form, then this is the same as saying it is a differential form of degree $k$.
So $df$ is a $1$-form, i.e. it has degree $1$ and $dx^\mu\wedge dx^\nu$ is a $2$-form, i.e. it has degree $2$.
Given a $1$-form, e.g. $M\,dx+N\,dy$ (in two dimensions), its exterior derivative is a $2$-form, $$\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)\,dx\wedge dy.$$
Here $k=1$ and $k+1=2$.
Edited in response to comments:
Perhaps this will make it more clear. Let's work in two dimensions again, to make things simple. When we say $df$ is a $1$-form, this isn't saying that the map $f\mapsto df$ is a $1$-form. Rather, for a particular function $f$, $df$ is itself a $1$-form, that is, a function from vectors to scalars.
Explicitly, $$df=\frac{\partial f}{\partial x}\,dx+\frac{\partial f}{\partial y}\,dy.$$
This is itself a covector which maps a vector to a scalar,
$$A\frac{\partial}{\partial x}+B\frac{\partial}{\partial y}\mapsto A\frac{\partial f}{\partial x}+B\frac{\partial f}{\partial y}.$$
More generally, the $1$-form $M\,dx+N\,dy$ will map vectors to scalars by
$$A\frac\partial{\partial x}+B\frac\partial{\partial y}\mapsto AM+BN.$$
(You might recognize this as just the dot product of $(A,B)$ and $(M,N)$.)
How do $2$-forms work, then? In two dimensions, a general $2$-form just looks like $P\,dx\wedge dy$. This should be a multilinear map taking two vectors and giving a scalar. The corresponding map is
$$\left(A_1\frac\partial{\partial x}+B_1\frac\partial{\partial y},A_2\frac\partial{\partial x}+B_2\frac\partial{\partial y}\right)\mapsto (A_1B_2-A_2B_1)P.$$
The reason there are two terms is that differential forms correspond to antisymmetric tensors; you can think of $dx\wedge dy$ as corresponding to the $(0,2)$-tensor $dx\otimes dy-dy\otimes dx$, which should make it more clear where the above expression comes from.
This is the sense in which differential forms are multilinear maps.
The exterior derivative operator then takes a differential form of degree $k$ to a differential form of degree $k+1$. One example I gave above, where if $\omega=M\,dx+N\,dy$ is a $1$-form, then $d\omega=(N_x-M_y)\,dx\wedge dy$ is a $2$-form.
Another example is if $f$ is a scalar function, then that is in fact a $0$-form.
In that case, $df=f_x\,dx+f_y\,dy$ is a $1$-form.
