Exterior derivative = infinitesimal change in differential form? For simplicity I'll work in $M=\mathbf R^2$.
Given $f\in C^\infty(M)=\Omega^0(M)$, its exterior derivative $df$ is a 1-form that eats a tangent vector and spits out the best linear approximation of (the change in) $f$ if we walk along the direction specified by that vector. In other words, given a point $(x,y)\in M$ and a tangent vector $(dx,dy)\in T_{(x,y)}M$, our 1-form $df=\displaystyle\frac{\partial f}{\partial x}\,dx+\frac{\partial f}{\partial y}\,dy$ eats $(x,y,dx,dy)\in TM$ and spits out a real number that's supposed to be the infinitesimal change in $f$.
The part I never really wrapped my head around is the exterior derivative of higher forms. In coordinates, it's usually defined to be $d(g\,dx+h\,dy)=dg\wedge dx+dh\wedge dy$ (and analogously for higher forms). Question: Can I interpret this to be the "infinitesimal change in $\omega=g\,dx+h\,dy$"?
Thus, instead of thinking of $d:\Omega^k(M)\to\Omega^{k+1}(M)$, can I think of $d\omega$ as eating a tangent vector and spitting out the infinitesimal change in $\omega$? Paraphrased: Does $X\lrcorner\, d\omega$ represent the infinitesimal change in $\omega$ in the direction $X$?
I've noted that $X\lrcorner\,d\omega$ is "half" of Cartan's magic formula, which is also supposed to represent the infinitesimal change in $\omega$ if we flow along $X$, and this just completely confused me. At this point, I'm not even sure I know what I mean by "infinitesimal change in $\omega$" anymore. Is there any hope in trying to understand things the way I'm currently trying to, or should I just abandon this altogether and just live with the axioms?
 A: Think about the special case of vector calculus in $\Bbb R^3$, where the exterior derivative reduces to the gradient, curl and divergence. The latter two can't really be thought about in this manner - they don't capture the full first-order variation of a form, only the coordinate-invariant components that come into play in Stokes theorem.
The fact that we're throwing away some information is clear in the coordinate formula for the general case - by taking the antisymmetric part of the partial derivative we're discarding the entire symmetric part, so a form (other than a function or a form of top degree) can vary in a symmetric fashion yet still be closed.
The way I think about the exterior derivative is as "the operator that makes Stokes theorem true" - i.e. as something in the spirit of 
$$ d\omega(X_1, \ldots, X_n) = \lim_{r \searrow 0} \frac1{r^{n-1}}\int_{\partial \square(n,r)} \omega $$ where $\square(n,r)$ is a "cuboid" with "edges" generated by the vectors ${rX_1, \ldots, rX_n}$. In the infinitesimal picture this integral over the boundary ends up being a certain sum of finite differences across opposing faces, so you end up with a first-order differential operator.
I'm not aware of a useful interpretation that separates out the "new input vector" like you're looking for - this seems unlikely precisely due to the antisymmetrization, which mixes that "direction of differentiation" in with all the other vectors. You certainly can't express $X\lrcorner d\omega$ in terms of derivatives in the $X$ direction alone.
The idea of directional infinitesimal change in a form is better captured by taking a covariant derivative, but of course this requires the extra structure of a connection. This should be no surprise - after all, 1-forms are basically the same thing as vectors, and we need a connection to differentiate vectors in a familiar fashion.
