Let $M$ be a smooth manifold and $X,Y,Z \in \mathfrak{X}(M)$ be smooth vector fields on $M$. Then we have that $$\mathscr{L}_{X + Y} Z = \mathscr{L}_XZ + \mathscr{L}_Y Z$$ where $\mathscr{L}_XY$ denotes the Lie derivative of $Y$ along $X$.
The proof we had in the lecture uses the definition of the Lie derivative and the auxiliary function $$\Psi_t := \Phi^{X + Y}_{t} \circ\Phi^Y_{-t} \circ \Phi^X_{-t}$$ where $\Phi$ denotes the flow of the respective vector fields. Now the claim is that $$\Psi_0 = \operatorname{id}_M \qquad \text{and} \qquad \frac{d}{dt}\Big\vert_{t = 0} \Psi_t = 0$$ Whereas the first is immediate from $$\Psi_0 := \Phi^{X + Y}_{0} \circ\Phi^Y_{0} \circ \Phi^X_{0} = \operatorname{id}_M \circ\operatorname{id}_M \circ \operatorname{id}_M = \operatorname{id}_M $$ by the properties of the respective flows, I have some trouble understanding why the second claim is true. Any help is appreciated.
Let $p\in M$ and define a map $f:(-\epsilon,\epsilon)^3\to M$ by$$(t_1,t_2,t_3)\mapsto\Phi_{t_3}^{X+Y}\circ\Phi_{t_2}^Y\circ\Phi_{t_1}^X(p).$$ It follows immediately from the construction that$$\frac{\partial f}{\partial t_1}(0,0,0)=X(p),\quad\frac{\partial f}{\partial t_2}(0,0,0)=Y(p),\quad\frac{\partial f}{\partial t_3}(0,0,0)=X(p)+Y(p).$$The desired claim follows now from applying the chain rule to the composition$$t\mapsto(-t,-t,t)\mapsto f(-t,-t,t).$$