How would you prove the identity
$\displaystyle \frac{\partial}{\partial s}\Psi_{s^*} \mathbb{X} = (-1)L_{\mathbb{Y}}\Psi_{s^*}\mathbb{X}$
where $\Psi_{s}$ is the flow of $\mathbb{Y}$ and $\Psi_{s^*}$ is its push forward. $\mathbb{X}$ and $\mathbb{Y}$ are vector fields.
The Lie derivative with respect to $\mathbb{X}$ is defined by $L_{\mathbb{X}}\mathbb{Y} = [\mathbb{X},\mathbb{Y}]$
The Lie bracket is defined by $\displaystyle \frac{\partial}{\partial s}\Bigg|_{s=0} \Psi_{s^*}\mathbb{X} = [\mathbb{X}, \mathbb{Y}]$
I think one of main issues is how to rid of the $\Bigg|_{s=0}$ and make it more general.
We want to show that
$\frac{\partial}{\partial s} \Psi_{s^*} X = [\Psi_{s^*}X, Y]$
(using the definition of $L$ and antisymmetry of $[,]$), i.e. that
$\frac{\partial}{\partial s} \Psi_{s^*} X = \left. \frac{\partial}{\partial t} \right|_{t=0} \Psi_{t^*} \left( \Psi_{s^*} X\right)$.
But we can write the RHS as
$\left. \frac{\partial}{\partial t} \right|_{t=0} \Psi_{t+s^*} X$
and this is just
$\left. \frac{\partial}{\partial t} \right|_{t=s} \Psi_{t^*} X$,
which is precisely what we mean on the LHS.