classical mechanics - Hamilton-Jacobi formalism and on-shell actions

My question is essentially how to extract the canonical momentum out of an on-shell action.

The Hamilton-Jacobi formalism tells us that Hamilton's principal function is the on-shell action, which depends only on the coordinates $q$. Therefore, if $S$ refers to the off-shell action with Lagrangian $L$, a result for the canonical momentum is

$$\frac{\partial L}{\partial \dot q} = p = \frac{\partial S^\text{on-shell}}{\partial q}$$

In theory, when evaluating the on-shell action, one plugs in the EOM into $S$ and all that's left are boundary terms. If we illustrate this with the harmonic oscillator in 1D, one gets

$$\begin{align} S^\text{on-shell} &= \int_{t_i}^{t_f} L dt \\ &= \frac{m}{2} \int_{t_i}^{t_f} \left( \dot q^2 - \omega^2 q^2 \right) dt\\ &= \frac{m}{2} \int_{t_i}^{t_f} \left( \frac{d}{dt} (q \dot q) - q \ddot q - \omega^2 q^2 \right) \\ &= \frac{m}{2} \left[ q(t_f) \dot q(t_f) - q(t_i) \dot q(t_i) \right] \end{align}$$

Now, this depends on $\dot q$, whereas it should depend only on generalized coordinates. Also, the on-shell action depends on the boundary values of our coordinate $q$, so we can't differentiate w.r.t. $q(t)$ to get a conjugate momentum $p(t)$ for all times, like it is usually done in the Lagrangian formalism.

What am I not seeing?