Consider two smooth vector fields $X,Y$, denote $[X,Y]$ their Lie bracket, $e^{tX}$ (resp. $e^{tY}$) the flow of $X$ (resp. $Y$) and $\varphi * X$ is the pushforward of $X$ by a diffeomorphism $\varphi$.
I would like to show: $$ \frac{d }{dt}(e^{-tX}* Y) = e^{-tX} * [X,Y] $$
using only definitions of the flow, pushforward and Lie brackets (no use of BCH...)
Observe that $[X,Y] = \frac{d}{dt}_{\mid t=0} e^{-tX}*Y$.
and
$$ e^{-tX}* Y = \frac{d }{ds}_{\mid s=0} e^{-tX}\circ e^{sY} \circ e^{tX} $$
I would double check the question that's being asked since there seems to be a minus sign error. You should actually have
$$ \left . \frac{d}{dt} \right |_{t=0} e^{-tX}*Y \;\; =\;\; -[X,Y]. $$
You get this by the following: \begin{eqnarray*} \left . \frac{d}{dt} \right |_{t=0} e^{-tX}*Y & = & \left . \frac{d}{dt} \right |_{t=0} \left . \frac{d}{ds} \right |_{s=0} e^{-tX} e^{sY} e^{tX} \\ & = & \left . \frac{d}{dt} \right |_{t=0} \left (Ad_{e^{-tX}}\right )_*Y \\ & = & -XY + YX \;\; =\;\; -[X,Y]. \end{eqnarray*}
From this we can instead prove that $$ \frac{d}{dt} e^{-tX}*Y \;\; =\;\; e^{-tX}*\left (-[X,Y]\right ). $$
Observe:
\begin{eqnarray*} \frac{d}{dt} e^{-tX}*Y & = & \frac{d}{dt} \left . \frac{d}{ds} \right |_{s=0} e^{-tX}e^{sY}e^{tX} \\ & = & \frac{d}{dt}e^{-tX}Ye^{tX} \\ & = & \left (Ad_{e^{-tX}}\right )(-XY + YX) \\ & = & \left (Ad_{e^{-tX}}\right ) (-[X,Y]) \\ & = & \left . \frac{d}{ds} \right |_{s=0} e^{-tX} e^{-s[X,Y]}e^{tX} \\ & = & e^{-tX} * (-[X,Y]). \end{eqnarray*}