If the idea behind the exterior derivative $d$ is that it tells us how quickly a $k$-form changes along every possible direction, why is $d(xdx)=0$ even though $xdx$ varies with $x$?
I understand the mechanics of exterior calculus but I'm trying to understand the intuition behind why it's the right tool for "differentiating" vector fields, rather than say working with vector-differences like a standard derivative (e.g. $\frac {\partial\overrightarrow v}{\partial x}=\lim\limits_{h \to 0} \frac{\overrightarrow v(x+h,y)-\overrightarrow v(x,y)}{h}$).
Consider a change of variables, $y = \frac12 x^2$, then $\frac{\mathrm dx}{\mathrm dy} = \frac 1x$. This implies $x\, \mathrm dx = \mathrm dy$, which is "constant"! This means that when we use different coordinates, we will disagree about what is "constant" for differential forms. We don't have this problem with functions: if $f$ is constant then no matter how you change variables the result is constant.
Therefore we have to accept that constancy is not a "real" concept for differential forms (unless you install additional structure called "connections" — just like how velocity is relative unless you designate a fixed body which we compare velocity to), and $\mathrm d(x\,\mathrm d x) = 0$ shouldn't be rejected solely because it is not constant from your viewpoint. In fact your first sentence is almost true: The exterior derivative shows all the "observable" rate of change of a $k$-form, and $x\,\mathrm dx$ only changes in a non-observable way.