Let $\frak{g}$ be a finite-dimensional real Lie algebra, $\varphi: \bigotimes^l \frak{g} \to \mathbb{R}$ a symmetric multilinear functional, and $\psi \in \Omega^k(M; \bigotimes^l \frak{g})$ a $\bigotimes^l \frak{g}$-valued form on a smooth manifold $M$. Then the composite $\varphi \circ \psi \in \Omega^k(M)$ is a $k$-form on $M$ (this is true even if $\varphi$ is not symmetric).
In the famous Chern-Simons 1974 paper on Characteristic Forms and Geometric Invariants, it is stated (without proof) that in this situation $d(\varphi \circ \psi) = \varphi\circ d\psi$ where $d$ is the exterior derivative. Is there a coordinate-free proof for this fact? Also, is the symmetry of the multilinear map $\varphi$ a necessary requirement, or would this still be true even if $\varphi$ is merely multilinear?
Note that the Lie algebra structure of $\frak{g}$ is most likely unimportant for this statement.
I found an answer to my question in Dupont's Fibre Bundles and Chern-Weil Theory [PDF]. Given a vector space valued form $\omega: \bigwedge^k TM \to V$, the exterior derivative commutes with post-composition by any linear map $\varphi: V \to W$. That is, $d(\varphi \circ \omega) = \varphi \circ d \omega$. So the symmetry of the map to my question was not relevant; it was only necessary so that the resulting form would be closed.