Let $M$ be a smooth manifold with a Poisson tensor $\pi$.
Let $X_f = \pi^{\#}(df)$, $X_g=\pi^{\#}(dg)$ be two Hamiltonian vector fields.
Let $\Phi_g^u:M\to M$ be the time-$u$ flow of $X_g$.
(That is, $\Phi_g^u(x)$ is the result of integrating $X_g$ for time $u$ starting from the point $x$.)
Claim: The pushforward of $X_f$ by $\Phi_g^{-u}$ is Hamiltonian in the following way: $(\Phi_g^{-u})_*X_f=X_{f \circ \Phi_g^u}$.
Attempt: Let $Y = (\Phi_g^{-u})_*X_f$ and let $\Phi_Y^t$ denote its time-$t$ flow.
We know that $\Phi_f^t \circ \Phi_g^{u} = \Phi_g^u \circ \Phi_Y^t$ by pushforward, hence $\Phi_Y^t = \Phi_g^{-u} \circ \Phi_f^t \circ \Phi_g^{u}$.
Also $X_{f \circ \Phi_g^u} = \pi^{\#}(d(f \circ \Phi_g^u)) = \pi^{\#}(df \circ d\Phi_g^u)$.
Now I just need to take a derivative and/or view $\Phi_g^{\pm u}$ as a change of coordinates, but how exactly?
The time-$u$ flow $\Phi_{g}^{u}$ of $X_{g}$ is a Poisson diffeomorphism, i.e. $(\Phi_{g}^{u})_{*}\pi=\pi$. Expressing this in terms of the sharp map gives $$ \pi^{\sharp}=(\Phi_{g}^{u})_{*}\circ\pi^{\sharp}\circ(\Phi_{g}^{u})^{*}. $$ In particular, we get $$ (\Phi_{g}^{u})_{*}X_{f\circ\Phi_{g}^{u}}=(\Phi_{g}^{u})_{*}X_{(\Phi_{g}^{u})^{*}f}=(\Phi_{g}^{u})_{*}\big(\pi^{\sharp}(d(\Phi_{g}^{u})^{*}f)\big)=(\Phi_{g}^{u})_{*}\big(\pi^{\sharp}((\Phi_{g}^{u})^{*}df)\big)=\pi^{\sharp}(df)=X_{f}. $$ Composing both sides with $(\Phi_{g}^{-u})_{*}$ gives the desired equality $$ X_{f\circ\Phi_{g}^{u}}=(\Phi_{g}^{-u})_{*}X_{f}. $$