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It is easy to show that the differential forms of order $1$ obeys a form of chain rule. To be precise, $d(f(g(x)) = f^\prime(x) d(g(x))$. This can be for example proved by fixing a co-ordinate basis $x_i$ for the tangent space at some point, and looking at the action of $df$ on each $\partial/\partial x_i$.

My first question is whether there exists a co-ordinate free proof of this.

The more important question is whether some such rule exists for higher order differential forms. The motivation is my attempt to grasp how similar differential forms and the usual process of differentiation are.

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The composition $f \circ g$ can be viewed as the pullback $g^*f$. Your equation says that $d(g^*f) = g^*df$, i.e. that the exterior derivative $d$ commutes with pullbacks. The same is true for higher-order forms: $d(g^*\omega) = g^*d\omega$ for any differential form $\omega$.

Here's a coordinate-free proof of $d(g^*f) = g^*df$. If $X$ is a tangent vector at a point $p$, we have

$$d(g^*f)_p(X) = X(g^*f) = (g_*X) f = df_{g(p)}(g_*X) = (g^*df)_p(X).$$

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Answer to the first question: $d_x(f \circ g)$ is just the differential of the smooth function $f \circ g$ at $x$, so by the chain rule for differentials it equals $d_{g(x)} f \circ d_x g$. But $d_{g(x)} f$ is just multiplication by $f'(g(x))$, so $d_x (f \circ g) = f'(g(x)) d_x g$.

For your second question to make sense you have to be more specific what $f$ should correspond to for higher order differential forms.

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