Quote:
"Let $X_1,Y_1$ and $X_2,Y_x$ be two pairs of vectors fields compatible by $\Phi$. Then $[X_1,X_2]$ and $[Y_1,Y_2]$ are compatible by $\Psi$. Consider $P\in M$ and $Q=\Phi(P)$, and let $g$ be a $C^2$ function in a neighborhood of $Q$. Then $ [Y_1,Y_2](g) =Y_1[Y_2(g)]-Y_2(Y_1(g)) =X_1[X_2(g\circ \Phi)]-X_2(X_1(g\circ \Phi)) =[X_1,X_2](g\circ \Phi) $."
During the step $Y_1[Y_2(g)]-Y_2(Y_1(g)) =X_1[X_2(g\circ \Phi)]-X_2(X_1(g\circ \Phi))$, looking at the term $Y_1[Y_2(g)] =X_1[X_2(g\circ \Phi)]$. By the definition of compatible, $[(Y_{2})_Q](g) =[(Y_2)_{\Phi(P)}](g) =[(\Phi_*)_PX(P)](g) =X_P(g\circ\Phi)$ That I could understand. However, how did $Y_1\rightarrow X_1$ in the expression?
Suppose $X$ and $Y$ are vector fields compatible by $\Phi$, i.e. $Y\circ \Phi = T\Phi\circ X$, or equivalently $Y_{\Phi(P)} = T_P\Phi(X_P)$ for all $P$. Then $Y_{\Phi(P)}g = T_P\Phi(X_P)(g) = X_P(g\circ \Phi)$, or in other words, $$Y(g)\circ\Phi = X(g\circ\Phi).$$ Applying this twice $$ Y_1(Y_2(g))\circ \Phi = X_1(Y_2(g)\circ\Phi) = X_1(X_2(g\circ\Phi)), $$ and so $$ [Y_1,Y_2](g)\circ\Phi = [X_1,X_2](g\circ\Phi). $$ If you're being sloppy, you can leave out the $\Phi$ on the left hand side, as done in your quote.