Prove that the Lie derivative of a vector field $Y$ in the direction of $X$ is the commutator $[X,Y]$ I'm trying to follow a proof that the Lie derivative of a vector field $Y$ in the direction of a vector field $X$ is the commutator $[X,Y]$. We write $\Phi_t$ for the flow of $X$.
How do I get from $$\frac{d}{dt}\vert _{t=0}\Phi^*_t(Y(\Phi^*_{-t}(f)))$$ to
$$\frac{d}{dt}\vert _{t=0} (\Phi^*_t(Y(f))-Y(\Phi^*_t(f))$$ ? I'm sure it's simple but I'm just not seeing it. 

Formally, this is because of the product rule (e.g. $\frac{d}{dt}\vert_{t=0} (A(t)BC(t)) = A'(0) B C(0) + A(0) B C'(0)$):
$$
\frac{d}{dt}\vert_{t=0} \Phi_t^*(Y(\Phi_{-t}^* (f))) = (\frac{d}{dt}\vert_{t=0} \Phi_t^*) (Y(\Phi_0^* f)) + \Phi_0^* Y(\frac{d}{dt}\vert_{t=0} \Phi_{-t}^* f).
$$
Now, use the fact that since $\Phi_t$ is a flow, $\Phi_0$ is the identity map and for the second term make the substitution $t \mapsto -t$.  The $\frac{d}{dt}\vert_{t=0}$ comes out in the second term since $Y$ is a linear operator and independent of $t$.
To convince yourself that you can apply the product rule to these pullbacks of diffeomorphisms you can either prove it using the same proof of the product rule for functions or you can observe that locally this is all just matrix multiplication (with functions as entries) and that the product rule holds for matrix multiplication of function valued matrices.
