What is the exterior derivative intuitively?

Many years back I wrote something about an intuitive way to look at differential forms here. In particular, figure 4 illustrates Stokes' theorem in a way that generalises to higher dimensions. Note that these are just sketches for intuition, and I've found them useful for illustrating various fields arising in physics, but they're not anything rigorous. They're also, in some sense, dual to the diagrams in Misner, Thorne and Wheeler. (There are some errors in my document, but I lost the source code many years ago...)

I think that the best explanation is in Arnold's book "Mathematical methods of classical mechanics". Here it is: after fixing a chart on a manifold one can say that the value of $d\omega$ ($\omega$ is a n-form) on tangent vectors $(\xi_1, ...,\xi_{n+1})$ at point $x_0$ equals to the coefficient of the $(n+1)$-linear part of the function $F(\varepsilon)=\int_{\partial V(\varepsilon)} \omega$, where $V(\varepsilon)$ is a "curvilinear parallelepiped" with vertexes $x_0, x_0+\varepsilon \xi_1, ..., x_0+\varepsilon \xi_{n+1}$: $F(\varepsilon)=(d\omega)(x_0)(\xi_1, ...,\xi_{n+1})\varepsilon^{n+1}+o(\varepsilon^{n+1})$.

For 1-forms, you can get some intuition for exterior differentiation from how it shows up in Frobenius's theorem which states that a distribution D is integrable if and only if the ideal of differential forms that are annihilated by it is closed under exterior differentiation:

Let $\alpha$ be a 1-form on $M$. If $\alpha$ does not vanish, then ker $\alpha_x$ is a hyperplane in the tangent space to $M$ at $x$. Thus ker $\alpha$ is a hyperplane field in $TM$ (and is an example of a distribution). At every point in M, you should visualize a hyperplane passing through that point.

Frobenius's theorem gives conditions on whether this hyperplane field is integrable, that is, if one can fit the planes together to form a foliation by hypersurfaces in $M$. For a hyperplane field defined by a single 1-form one can fit the planes together if and only if $d\alpha$ mod $\alpha$ is zero. This is usually expressed by the vanishing of $\alpha\wedge d\alpha$.

(In the general case, where instead of $\alpha$ we have a set of linearly independent 1-forms $\{\alpha_j\}_{j=1}^r$, the ideal in the algebra of differential forms on $M$ generated by $\{\alpha_j\}_{j=1}^r$ must be closed under exterior differentiation; equivalently $d\alpha_j\wedge\alpha_1\wedge\cdots\wedge\alpha_r=0$ for all $j$).

Two simple examples:

(1) if $\alpha=df$ then the field of hyperplanes ker $\alpha$ is actually tangent to the hypersurfaces $f=$const (and of course $d\alpha=0$).

(2) If $\alpha = g df$ for some non-vanishing function $g$, e.g. $\alpha=ydx$ in the upper half plane of $\mathbb{R}^2$, then this is just as good, since ker $\alpha$ is still tangent to $f=$const. Note that $d\alpha=dg\wedge df=(dg/g)\wedge\alpha$, which vanishes mod $\alpha$ and thus $\alpha\wedge d\alpha=0$.

Hence $\alpha\wedge d\alpha$, or $d\alpha$ mod $\alpha$ roughly measures how far this hyperplane field defined by ker $\alpha$ is from being tangent to hypersurfaces.

(I got the ideas from Appendix B of Ivey and Landsberg's book Cartan for Beginners. Thanks to Marcos Cossarini and Ben McKay for pointing out in the comments that the original version of this was wrong!)

Here's an example of a hyperplane field which is not tangent to any hypersurfaces. $\alpha = dz-y dx$ on $\mathbb R^3$ and $\alpha\wedge d\alpha = dz\wedge dx \wedge dy$: