I hope I am right.

Fix $t \ge 0$. For each $\omega$, $f(\cdot,\omega)$ is square-integrable hence integrable on $[0,t]$

For every $t$, the map $(s,\omega) \mapsto f(s,\omega)$ is measurable on $[0,t] \times \Omega$.

So $\omega \mapsto \int_0^t f(s,\omega)$ is measurable. This result is contained in Fubini theorem for integrable functions in both variables. Here, one may apply this result first with $f_n = f 1_{[|f| \le n}$ and apply Lebesgue dominated convergence theorem to get the pointwise convergence of the integrals of $f_n$ to the integral of $f$ and deduce that $\int_0^t f(s,\omega)$ depends measurably on $\omega$.

Therefore, the process $\int_0^\cdot f(s,\cdot)$ is adapted. Since it is continuous, it is progressively measurable.

Partial answer

The question may be whether the process $f$ is progressively measurable under the assumptions. If yes, we can conclude as follows.

Fix $t \ge 0$. For each $\omega$, $f(\cdot,\omega)$ is square-integrable hence integrable on $[0,t]$

For every $t$, the map $(s,\omega) \mapsto f(s,\omega)$ is measurable on $[0,t] \times \Omega$.

So $\omega \mapsto \int_0^t f(s,\omega)$ is measurable. This result is contained in Fubini theorem for integrable functions in both variables. Here, one may apply this result first with $f_n = f 1_{[|f| \le n}$ and apply Lebesgue dominated convergence theorem to get the pointwise convergence of the integrals of $f_n$ to the integral of $f$ and deduce that $\int_0^t f(s,\omega)$ depends measurably on $\omega$.

Therefore, the process $\int_0^\cdot f(s,\cdot)$ is adapted. And since it is continuous, it is progressively measurable.