Given a pair of autonomous Hamiltonian vector fields $X_H,X_K\in\mathfrak{X}(M)$, with flow maps which are respectively $\Phi^t,\Psi^s$, is $\Phi^t\circ \Psi^s$ the $(t+s)-$flow map of some Hamiltonian system? I think this is the case, possibly with the final map being the flow of a time-dependent Hamiltonian system.
I know that the composition map is a symplectomorphism. However, since not all the functions in this space are flows of Hamiltonian systems, I am not sure of the answer to this question.
Here is an additional explanation of what I am interested in.
I do not mean that for any $t$ and $s$, the Hamiltonian of the resulting vector field remains unchanged. More precisely, here are some settings which, for my interest, are not counterexamples: $$ \Phi^t\circ \Psi^s = \varphi_{X_L}^{t+s} $$ $$ \Psi^t\circ \Phi^s = \alpha_{X_R}^{t+s} $$ $$ \Phi^u\circ \Psi^v = \varphi_{X_W}^{u+v}, $$ where $L,R,W$ are three Hamiltonian functions which possibly do not coincide.
This means that, when we fix $t$ and $s$, there is a Hamiltonian vector field $X_{U}$ so that its $(t+s)-$flow is the same map as $\Phi^t\circ \Psi^s$.
Firstly, the length of the parametrizing time interval does not matter: if $\phi_t$ is the flow of $X_t$ for $t \in [0, T]$, i.e. $\frac{d}{dt} \phi_t(x)=X_t(\phi_t(x))$ then defining $\psi_s=\phi_{Ts}$ with $s\in [0,1]$ one has $\frac{d}{ds} \psi_s(x)=TX_{Ts}(\phi_{Ts}(x))$. This says that the reparameterized $\psi_s$ is generated by $Y_s=TX_{Ts}$. Of course if $X_t$ is Hamiltonian, then so is $Y_s$ and vice versa.
Thus we can assume that both $\Phi$ and $\Psi$ are time 1 maps of Hamiltonian systems, i.e. $\Phi=\phi_1$ for a family $\phi_t$ with $t\in[0,1]$, and $\Psi=\psi_1$ for a family $\psi_t$ with $t\in[0,1]$. Denote the corresponding Hamiltonian vector fields by $X_t$ and $Y_t$. Now, consider
$$\theta_t=\phi_t\cdot \psi_t$$ so that
$$\Theta=\theta_1=\phi_1\cdot \psi_1=\Phi\cdot\Psi.$$
We just need to show that $\theta_t$ is a family of Hamiltonian diffeos. Of course this means we need to compute the generating (time dependent) vector field and see that it's Hamiltonia.
To do this we differentiate
$$\frac{d}{dt}\theta_t(x)=\frac{d}{dt}(\phi_t\cdot \psi_t(x))=$$ $$ (\frac{d}{dt}\phi_t)(\psi_t(x))+(\phi_t)_*(\frac{d}{dt}\psi_t(x))=$$ $$X_t(\phi_t(\psi_t(x)))+(\phi_t)_*(Y_t(\psi_t(x)))=X_t(\theta_t(x))+(\phi_t)_*Y_t(\phi_t^{-1}\theta_t(x))$$
so $\theta_t$ is generated by the vector field $Z_t(x)=X_t(x)+(\phi_t)_*Y_t(\phi_t^{-1}(x))$.
Now all that remains is to show that this is Hamiltonian. What is the corresponding function? Well, if the function for $X_t$ is $H_t$ and for $Y_t$ is $K_t$, then I claim that for $Z_t$ it's $H_t+K_t\cdot\phi_t^{-1}$. Of course it's the $K_t\cdot\phi_t^{-1}$ part that's the issue. Naturality of Hamiltonian vector fields under symplectic diffeos means that for any function $K$ and any symplectomorphism $f$, we have $ f_*(X_{K\cdot f}(p))=X_K(f(p))$. Applied to $f=\phi^{-1}$ this gives what we want.
As a remark, it is in fact the case that we need time-dependent flows. In fact, it follows from the simplicity of the group of Hamiltonian diffeos that every Hamiltonian diffeomorphism is a product of autonomous ones (it's a nice exercise to check that inverse of an autonomous Hamiltonean diffeo is also autonomous, so the set of Hamiltonean diffeos that are a finite product of autonomous ones forms a conjugation invariant subgroup, hence is the whole group), but sometimes you need arbitrarily many of them (see https://arxiv.org/abs/1207.0624 and follow-up work).