Let $A,B \in End(\mathbb{R}^n)$. Let $\overline{A}$ denote the vector field $x \mapsto Ax$, and $\overline{B}$ likewise.
In An Introduction to Differential Manifolds, Lafontiane claims that $[\overline{A}, \overline{B}]_x = (BA - AB) \cdot x$ (where the bracket is of the derivations associated with the vector field). I understand that it is possible to do this using coordinates, but this gets extremely messy. Is there some coordinate-free way to see this?
You could use the identity $[X,Y]=\mathcal{L}_XY$.
The flow of the vector field $\overline{A}$ is the one parameter family of linear isomorphisms $$a_t(x)=e^{tA}x.$$As $a_t$ is linear for every $t$, the differential of $a_t$ is again $a_t$. Hence,$$[\overline{A},\overline{B}]_x=\mathcal{L}_{\overline{A}}\overline{B}_x=\left.\frac{d}{dt}\right|_{t=0}\left(da_{-t}\left(\overline{B}(a_t(x))\right)\right)=\left.\frac{d}{dt}\right|_{t=0}e^{-tA}Be^{tA}x.$$The desired equality follows from the Leibniz rule.