Confusion on Wikipedia Intuition on Differential Forms In the Wikipedia article on the exterior derivative, the following intuition is given:

If a differential $k$-form is thought of as measuring the flux through an infinitesimal $k$-parallelotope at each point of the manifold, then its exterior derivative can be thought of as measuring the net flux through the boundary of a $(k + 1)$-parallelotope at each point.

I suppose I really just don't get this intuition at all, not even in the $k = 0$ or $k = 1$ cases. For one thing, what are we measuring the flux of? I wasn't even aware that the flux through a line segment was a well-defined notion. Also, what $k$-parallelotope are we measuring the flux through, and how exactly does a differential form encode this information?
So yeah, what's the deal with this explanation?
(A few extra points:

*

*I'd like to clarify that this question is not asking for a general intuition on differential forms. This question is about the Wikipedia intuition in particular, not any others.

*The textbook I'm working out of is Spivak's Calculus on Manifolds. The way the exterior derivative of a $k$-form is defined there is the following:
\begin{align}
\omega &= \sum_I \omega_I dx^I, \\
d\omega &= \sum_I d\omega_I \wedge dx^I, 
\end{align}
where $I$ is a multi-index from $1$ to $n$, with $k$ elements. Not sure if that helps, just thought I'd mention my perspective in case it's helpful.)

 A: Let's work in $\Bbb R^n$ for simplicity (this is about intuition, after all).
Suppose $\omega = \sum_k\omega_k\mathrm d x^k$ is a nice $1$-form (covector field) on $\Bbb R^n$: then it can be converted into a vector field $\boldsymbol\omega := \omega^\sharp = \sum_k\omega_k \partial_{x^k}$ with the same components[1]. When you evaluate $\omega(\boldsymbol v)$ for some vector field $\boldsymbol v$ on $\Bbb R^3$, you obtain a scalar field; the meaning of this scalar field is clarified by "translating" $\omega$ into $\boldsymbol \omega$: with the Einstein summation convention,
$$\omega (\boldsymbol v)=\omega_i\ \mathrm dx^i\left(v^j \frac{\partial}{\partial x^j}\right) = \omega_iv^i\equiv \boldsymbol \omega \cdot\boldsymbol v,$$
which is the scalar field obtained by projecting $\boldsymbol v$ onto $\boldsymbol \omega$ (or viceversa!) point by point. Thus, if you draw a small segment (a small $1$-parallelotope) at a specific $p\in\Bbb R^n$ in such a way that it is parallel to $\boldsymbol v$, in some sense, $\omega(\boldsymbol v)$ can be seen to represent a "(tangential) flux" along the segment. This is basically an infinitesimal interpretation of the line integral of $\boldsymbol \omega$ in vector calculus (ideally taken along a very short curve).
This intuition can be extended to $k$-degree forms, as long as you feed them $k$ vector fields instead of one; these fields then describe a $k$-parallelotope at all points.
The reasoning above is also why it is often said that $\omega$ are "made to be integrated": specifically, $k$-forms like to be integrated on manifolds of dimension $k$. If you imagine a regular enough $k$-manifold $M$ (embedded in $\Bbb R^n$, again for simplicity) and a $k$-form on it, then at each point $p\in M$ one could draw $k$ vectors $\boldsymbol v_{1p},\dots,\boldsymbol v_{kp}$ tangent to $M$, and if their modulus is taken to be smaller and smaller their $k$-parallelotope "approximates" $M$ better and better around $p$. As we know, the form $\omega$ describes a flux across every such parallelotope, and the integral over $M$ is a way to "compound" all these fluxes along $M$. In the case $k=1$, this represents the line integral from vector calculus (finite this time: think physical work of a force along a path); in the case $k=2$, this is a surface integral; and so on.
This digression on integrals comes in handy when thinking about the exterior derivative. If $\omega$ is a $k$-form, its exterior derivative $\mathrm d\omega$ is a $(k+1)$-form, and is naturally integrated along a regular $(k+1)$-manifold $M$. Stokes' theorem is the miraculous fact that, if this manifold has a boundary and is nice enough[2], then
$$\int_M \mathrm d\omega = \int_{\partial M}\omega. $$
(This is a generalization of the gradient, curl, and divergence theorems from vector calculus.) Now suppose you take $M$ and "contract it" around a point $p\in M$: then it looks more and more like a $(k+1)$-parallelotope at $p$, and the action of $\omega$ can again be interpreted as a flux of $\mathrm d\omega$ along this parallelotope. But by Stokes' theorem, this corresponds to the total flux of $\omega$ along the boundary faces of the parallelotope (which are $k$-parallelotopes)!

Footnotes. [1] This is not something we can do in all manifolds: the musical isomorphisms $(-)^\sharp$ and $(-)^\flat$ are only defined in manifolds where there is a (pseudo)metric tensor $g$, i.e. (pseudo-)Riemannian manifolds. Here, we are thinking of $\Bbb R^n$ as a Riemannian manifold with the standard Euclidean metric $g(\partial_{x^i},\partial_{x^k})=\delta_{ik}$.
[2] Since a $k$-form is alternating in its arguments, it matters in what order you feed it vector fields. Thus, when approximating a manifold $M$ by $k$-parallelotopes, you need to make sure that the parallelotopes "vary smoothly" along the surface (i.e. the vector fields $\boldsymbol v_i$ describing their edges must be smooth) and that, if you follow a closed loop from a point $p\in M$ to itself, you don't unknowingly switch some of the edges around. Not all manifolds can be "traced out" by vector fields with these properties; a famous counterexample is the Möbius strip (try it!). When the manifold does admit these edge fields, it is said to be orientable, and a choice of edge fields is said to define an orientation. Once you calculate an integral in a fixed orientation, switching two of the edge fields and calculating again leads to minus the integral you had before!
Stokes' theorem holds only for orientable manifolds with boundary, and relies on the fact that, once an orientation is chosen on $M$, it induces a natural orientation on $\partial M$—which is the one with respect to which the integral on the LHS needs to be calculated; otherwise, you get a pesky sign error.
