Let $M$ be an open set in $\mathbb{R}^d \times \mathbb{R}^d$ such that the flow of the Hamiltonian maps $M$ to itself and $G_t$ be the flow map of the Hamiltonian i.e. $G_t(x_0,p_0) = (x(t),p(t))$ if $x(0)=x_0$ and $p(0)=p_0$. Now we let $f_t:= f\circ G_t$. Let $\mathcal{A}=C^\infty(M)$.
Then we have $\frac{d}{dt}f_t = \{H,f_t\},$ where $\{\cdot, \cdot\}: \mathcal{A}\times \mathcal{A} \to \mathcal{A}$ is given by
$$\{f,g\}:=\sum_{i=1}^d (\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial x_i}-\frac{\partial f}{\partial x_i} \frac{\partial g}{\partial p_i}).$$
The proof proceeds as follows. Let $(x,p)\in M$ be given We denote by $(x(t),p(t))$ the solution of the Hamiltonian equation with initial data $(x,p)$. We compute, using the chain rule
$$\frac{d}{dt}f_t(x,p)=\frac{d}{dt}f(x(t),p(t))=\sum_{i=1}^d \big[ \frac{\partial f}{\partial x_i}(x(t),p(t))\dot{x_i}(t)+\frac{\partial f}{\partial p_i}(x(t),p(t))\dot{p_i}(t)\big].$$
Since $(x(t),p(t))$ solves Hamilton's equation the above expression is $$=\sum_{i=1}^d \big[\frac{\partial H}{\partial p_i} \frac{ \partial f}{\partial x_i}-\frac{\partial H}{\partial x_i}\frac{\partial f}{\partial p_i}\big](x(t),p(t)) = \{H,f_t\}(x,p).$$
In the last expression of the proof, I am not sure why $\frac{\partial f}{\partial x_i}(x(t),p(t))=\frac{\partial f_t}{\partial x_i}(x,p)$ and similarly for the partial derivative of $f_t$ with respect to $p_i$.
I am confused because $f_t=f\circ G_t$ so if we take the partial derivative of $f_t$ with respect to $x_i$ shouldn't we use the chain rule? From my understanding, the left side is $\frac{\partial f}{\partial x_i}(x(t),p(t))=\frac{\partial f}{\partial x_i}\circ G_t(x,p)?$ But how is this equal to $\frac{\partial f_t}{\partial x_i}(x,p)=\frac{\partial (f\circ G_t)}{\partial x_i}(x,p)?$ How do we just move $G_t$ inside the partial derivative?