Fundamental vector fields form a Lie-algebra homomorphism

For $p\in M$, write $a_p(g)=g\cdot p$. Since the identity differential of the exponential is the identity, $X_M(p)=-d(a_p{)}_{e}X$, so $X\mapsto X_M$ is linear. The curve ${\mathrm{\Phi }}_t(p)=\exp (-tX)\cdot p$ has velocity $X_M$ at every time, because $\exp (-(t+s)X)=\exp (-sX)\exp (-tX)$; hence $\mathrm{\Phi }$ is the global flow of $X_M$.

For fixed $g\in G$, [F4] gives $g\exp (-tY)g^{-1}=\exp (-t{\mathrm{Ad}}_{g}Y)$; differentiating this identity in its action on $g\cdot p$ gives $(g\cdot {)}_{\ast }{Y}_M=({\mathrm{Ad}}_{g}Y{)}_M$. Apply this with $g=\exp (tX)$, which acts as ${\mathrm{\Phi }}_{-t}$ by step 1.1, to obtain $({\mathrm{\Phi }}_{-t}{)}_{\ast }{Y}_M=({\mathrm{Ad}}_{\exp (tX)}Y{)}_M$.

By [F2], the derivative at $t=0$ of the left side in step 2.1 is ${\mathcal{L}}_{X_M}{Y}_M=[X_M,Y_M]$. By [F3] and linearity from step 1.1, the derivative of the right side is $({\mathrm{ad}}_{X}Y{)}_M=[X,Y{]}_M$. This proves the formula. The action need not be effective, free, or transitive; if either vector is zero or the group is zero-dimensional, both sides vanish. All flows used are global, so there is no endpoint issue. Countable choice is inherited exactly through [A1], [F1], [F3], and [F4].