Lie bracket and flows on manifold

I'm not sure if this is the answer you're looking for, since this is by computation. But it does not involve the chain rule, at least.

So... both sides of the equation are elements of the tangent space $T_pM$. To see that they are equal, we compute their action on a function $f:M \to \mathbb R$.

Now, by definition $$ [X,Y](f)(p) = \lim_{h \to 0} \frac 1h \left[ (Yf) \circ \phi_h^X(p)-Y(f)\right] - \lim_{h \to 0} \frac 1h \left[ (Xf) \circ \phi_h^Y(p)-X(f)\right] $$ The first term is equal to: $$ \lim_{h \to 0} \frac 1h \left[ (\lim_{k \to 0}\frac 1k [f \circ \phi_k^Y(p)-f(p)]) \circ \phi_h^X(p)-[\lim_{k \to 0}\frac 1k f \circ \phi_k^Y(p)-f(p) ]\right] $$ Setting $k=h$ (the functions are differentiable, so this shouldn't change the answer), we get $$ = \lim_{h \to 0} \frac{1}{h^2} \left[ f \circ \phi_h^Y \circ \phi_h^X (p) - f \circ \phi_h^X(p) -f \circ \phi_h^Y(p) + f(p) \right] $$ Doing the same for the other term, we get: $$ \lim_{h \to 0} \frac{1}{h^2} \left[ f \circ \phi_h^X \circ \phi_h^Y (p) - f \circ \phi_h^Y(p) -f \circ \phi_h^X(p) + f(p) \right] $$ Subtracting them, most terms cancel and we get $$ \lim_{h \to 0} \frac {1}{h^2}\left[ f \circ \phi_h^Y \circ \phi_h^X(p) - f \circ \phi_h^X \circ \phi_h^Y(p) \right] $$ Now we use that $(\phi_h^X)^{-1}=\phi_{-h}^X$ to get $$ \lim_{h \to 0} \frac {1}{h^2}\left[ f- f \circ \phi_h^X \circ \phi_h^Y \circ \phi_{-h}^X \circ \phi_{-h}^Y(p) \right] $$ But this is just the derivative of $\alpha$ (we traverse in the opposite direction, but that's okay)! Putting $t=h^2$ we get the result.

So all in all, this computation was not painfree, but it is clear why we need the square root signs.