Derivative of differential of time-dependent flow of a vector field

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I am reading the book "Morse Theory and Floer Homology" by Audin and Damian. And I am kind of stuck in understanding the last part of a proof in such book. Currently we are trying to prove

Proposition 5.4.5. If a critical point of $H$ is nondegenerate as a periodic solution of the Hamiltonian system, then it is nondegenerate as a critical point of the function $H$.
(Screenshot here)

So, part of the proof is here

Computed at the critical point $x$ of $H$, the first term is zero and the second one is exactly $$ \left(d^{2} H\right)_{x}\left(X_{f}(x), Z(x)\right)=\left(d^{2} H\right)_{x}(Y, Z) $$ We now suppose that $x$ is nondegenerate as a trajectory of $X_{H}$. This means that for every $Z \neq 0$, we have $$ T_{x} \psi^{1}(Z)-Z \neq 0. $$ (Screenshot here)

My main question is in page 138. So it says that $$\frac{d}{dt} T_x \psi^{t}(Z) = T_x \psi^{t} ([X_H, Z])$$

I do not understand where this equality comes from. I have been trying to see if this is a known property of time-dependent flows, but no success until now. Here $X_H$ denotes the time-dependent Hamiltonian vector field.

Here is the things I have tried until now:

  1. In the appendix of such book it is known that for time-dependent flows we have that $\frac{dY}{dt} = (dX_t)_{x(t)}Y$ iff $Y(t) = T_{x(t)}\psi^{t}(Y(0))$, where $X_t$ is any time-dependent vector field. So, we can apply this to our case where $X_t = X_H$, but after that I am not sure how to go from $(dX_H)_{x(t)}Y$ to $T_x \psi^{t}([X_H, Z])$.
  2. The other thing that I tried to imitate was the fact that for time-dependent vector fields we get a formula for $\frac{d}{dt} \psi^{*} Y_{t}$, where it is very similar with what I want, but at the end this formula is for the pullback of my flow $\psi$ and I want a formula for the push-forward.

Again, maybe this is a basic fact of time-dependent flows and it is easy to prove but if I could get any kind of hint or reference I could look to understand this I would be grateful.

Thanks in advance