I don't know if this answers directly your question, but based on your previous question, and what you said in this one, I thought it was worth it to write a few things. Tell me if this is completely off and I will delete this attempt to answer.
Having a zero Hamiltonian on-shell does not mean there are no transitions. For example, the free particle's Hamiltonian action is :
\begin{equation}
S=\int_0^1 dt \left( p_\mu\dot{x}{}^\mu-\frac{e}{2}(g^{\mu \nu}p_\mu p_\nu+m^2) \right) \tag{1}
\end{equation}
Where it is crystal clear that, on-shell, $\mathcal{H}=g_{\mu \nu}p_\mu p_\nu+m^2=0$. The ADM formulation of gravity, on its side, enjoys the Hamiltonian action:
\begin{equation}
S_G=\int_{\mathcal{M}} d^4x\left( \pi^{ij} \dot{\gamma}_{ij}-N\mathcal{H}_G-N_i \mathcal{P}^i \right) \tag{2}
\end{equation}
Where $\mathcal{H}_G$ is the Hamiltonian constraint and $\mathcal{P}^i$ is the diffeomorphism constraint.
Both $(1)$ and $(2)$ exhibit a null Hamiltonian, but we know that the free particle is evolving, so having this property does not imply there are no transitions.
One thing you can do is use the BFV formalism on the Hamiltonian path integral arising from $(2)$, as Qmechanic suggests in his answer to you previous question:
\begin{equation}
Z=\int\mathcal{D}\Phi_A \mathcal{D} \Phi^{\ast A}\,e^{i\int_{\mathcal{U}}d^4x (\Phi^{\ast A}\dot{\Phi}_A-\{Q,\psi\})}
\end{equation}
Where $\Phi$ are the fields, $\Phi^\ast$ the anti-fields, $Q$ the BRST charge of the theory, and $\psi$ a gauge fixing fermion you are free to choose as long as it is fermionic. Note that I used $\mathcal{U} \subset \mathcal{M}$, such that $\partial \mathcal{U}=f-i$, in your notations. You can follow https://journals.aps.org/prd/abstract/10.1103/PhysRevD.47.1420, "Microcanonical functional integral for the gravitational field" by Brown and York, and see that, upon ignoring the integral on the lapse function, you obtain indeed something of the form:
\begin{equation}
T_{i\rightarrow f}(N)={}_{\mathcal{P}}\langle f | e^{i\int_{\mathcal{U}}d^4x N\mathcal{H}_G} | i \rangle_{\mathcal{P}}
\end{equation}
Now, $N$ does not depend on the time parameter anymore. Moreover, the initial and final states satisfy the diffeomorphism constraint. So in fact, the problem seems to be that, contrarily to the free particle's Worldline action where reparametrizing the time parameter can be done without harm, you cannot repramatetrize the time. Indeed if you try to do so, you will not end up with one time interval, but with a field of time intervals:
\begin{equation}
I=\bigcup_{x\in i}[t_i(x),N(x)t_f(x)],\,\,\,\partial I \neq f-i
\end{equation}
Where $t_i(x)$ and $t_f(x)$ are the time position of the point $x$ in $i,f$. Of course, if you try to absorb $N$ in $t_f$, you end up with a final hypersurface $f$ which is not space-like anymore...
To conclude with your question: The gravitational path integral aims to describe a transition amplitude between two space-like hypersurfaces, but it is plagued with this reparametrization invariance and not by its null hamiltonian, and the fact that there is this very "problem of time" means that we actually don't quite know what to do. Of course, there are different approaches https://en.wikipedia.org/wiki/Problem_of_time#Proposed_solutions_to_the_problem_of_time, but there doesn't seem to be one solution that is accepted as consensus. There are transitions, but right now, no one can say "I have the right approach to quantum gravity, and it solves the problem of time". Remember that this is an entire field of research...
