Maybe this is a weird question...
It's known that the Lie derivative of one vector field with respect to another equals their Lie bracket. The proofs in the literature rely on viewing vector fields as derivations. I haven't found a "geometric proof" (where tangent spaces are models by equivalence classes of germs of curves).
Question. Where is this fact used in differential geometry? Why is it important?
On
One general reason is that it's often easy to compute. Given a complete vector field $X$, computing its flow $\Theta_{X}(t)$ generally amounts to solving a first order partial differential equation. The lie derivative of another vector field $Y$ is defined in terms of this flow as $\mathcal{L}_XY=\frac{d}{dt}\Theta_X^*(t)Y|_{t=0}$. The fact that $\mathcal{L}_XY=[X,Y]$ essentially states that "integrating" the PDE and differentiating w.r.t. the flow "cancel out" so that we can compute $\mathcal{L}_XY$ solely through simpler algebraic manipulations.
For a concrete case where this comes up, the study of Lie groups comes to mind. Flows (i.e. one dimensional Lie group actions) come up frequently here, and by differentiating them, Lie groups and their actions can be almost completely described in term of vector fields and their lie brackets (which form finite dimensional algebras).
Here's an application of interest in differentiable geometry, especially to physicists: since $£_XY$ is a Killing vector for KVs $X,\,Y$, KVs form a Lie algebra under the commutator we call a Lie bracket. The KVs on an $n$-dimensional manifold are known to (i) generate spacetime symmetries and (ii) form an at most $\frac12n(n+1)$-dimensional vector space, so there's a polynomial upper bound on the number of structure constants of this Lie bracket. These characterize gravity's Lie group as a gauge theory. Interestingly, the resulting group depends on not only the dimension of the manifold, but also its geometry.