I am reading some differential geometry notes and I stuck with understandig what seems to be a simple part of a proof but for me comes out of nowhere.
Let $M$ be a smooth manifold and let $X,Y$ be two smooth vector fields on the manifold. Then one can define the Lie derivative of $Y$ in direction of $X$ as
$$ L_X Y = \frac{d}{dt} ((\Psi_{-t})_* Y) \bigg \vert_0.$$
Here, $\Psi_{-t}$ denotes the local flow of the vector field $X$, that is in some neighborhood of some point one has $$ \frac{d}{dt} \Psi_t(q) = X_{\Psi_t(q)}.$$
$(\Psi_{-t})_*$ denotes the push-forward of $\Psi_{-t}$ and we have $$(\Psi_{-t})_* Y)_{\Psi_{-t}p} = {d\Psi_{-t}}_p Y_p$$ and thus $$(\Psi_{-t})_* Y)_{p} = {d\Psi_{-t}}_{\Psi_{t}(p)} Y_{\Psi_{t}(p)}.$$
Now in the proof that the definition above for the Lie derivative is the same as using $[X,Y]$ where this denotes the Lie bracket, the first step is as follows: For a function $f \in C^{\infty}(M)$, we have
$$ {L_X Y}_p \cdot f = \frac{d}{dt}(Y_{\Psi_{t}(p)} \cdot (f \circ \Psi_{-t}) ) \bigg \vert_{t = 0}.$$
This is already the step that I do not understand: From above we have
$$ {L_X Y}_p \cdot f = \frac{d}{dt}{d\Psi_{-t}}_{\Psi_{t}(p)} Y_{\Psi_{t}(p)}\bigg \vert_{t = 0} \cdot f.$$
How can we now get the function $f$ "inside" of the time derivative, and where does the change of the composition order come frome?
If the element in brackets was some $t \to M$ then maybe it would make more sense to me, since the elements of the tangent space $T_pM$ can be represented by such curves (or rather there derivatives at point $p$), but here we have a curve in $T_pM$.
Could anyone please explain this step a bit to me?
Note $$L_{X}Y(p)f = \frac{d}{dt}(d\psi_{-t}(\psi_t(p))Y(\psi_t(p))|_{t = 0})f =: (\frac{d}{dt}X_t(p)|_{t = 0})f$$
Note that $X_{t}(p) \in T_pM$ for all $t$ and using coordinates we have $$\frac{d}{dt}(X_t(p)f) = \frac{d}{dt}\sum_{i}\partial_if(p)X_{t}^i(p) = \sum_{i}\partial_if(p)\frac{d}{dt}X_t^i(p) = (\frac{d}{dt}X_t(p))f.$$ Hence $$\frac{d}{dt}(d\psi_{-t}(\psi_t(p))Y(\psi_t(p))|_{t = 0})f = \frac{d}{dt}(d\psi_{-t}(\psi_t(p))Y(\psi_t(p))f|_{t = 0}).$$
Now note that, by definition, $X_t(p)f = df(p)X_t(p)$. Hence by the chain rule $$d\psi_{-t}(\psi_t(p))Y(\psi_t(p))f = df(p)d\psi_{-t}(\psi_t(p))Y(\psi_t(p)) = d(f \circ \psi_{-t})(\psi_t(p))Y(\psi_t(p)) = Y(\psi_t(p))(f \circ \psi_{-t}).$$ This completes their proof.