First, your notation is nonstandard; when we write $\mathrm{Col}(\kappa,\lambda)$, this typically means the set of partial functions $p : \kappa \to \lambda$ of size $<\kappa$, i.e. reverse of yours.

First note the following fact:

> (1) If $\kappa$ is regular and $\mathbb P$ is $\kappa$-c.c., then every club subset of $\kappa$ in $V^\mathbb{P}$ contains a club from $V$.

To answer, "What is $\mathcal F_{\mathrm{Col}(\omega,\omega_1)}$?":  It is just the club filter on $\omega_2$.  Since this forcing has size $<\omega_2$, every new club contains a ground model club, so the new club filter is generated by the old.

For the next question, we use the following well-known result due to Harvey Friedman (generalized by Stavi and Abraham-Shelah):

> (2) If $S \subseteq \omega_1$ is stationary, then the forcing $\mathbb C(S)$ consisting of closed bounded subsets of $S$ ordered by end extension adds a club through $S$ without adding reals.

Let $S = S^{\omega_2}_{\omega_1}$, i.e. the ordinals of uncountable cofinality below $\omega_2$.  Then $\mathrm{Col}(\omega,\omega_1) * \dot{\mathbb C}(\check S)$ forces $S$ is in the new club filter on $\omega_2^V$.  By (1), $S$ remains stationary, so we may apply (2).

For the next question, the above argument shows that your ideal $\mathcal I$ is simply the nonstationary ideal on $\omega_2$.

I don't know much about forcing without choice, so I'll leave the rest to someone else.