I'm struggling with trying to find a formula for $E[f(X, Y)|\mathscr{F}]$ where $X$ and $Y$ are like in the title, $f$ is borel, and $E|f(X, Y)|<\infty$. I know that there exists a formula for $E[f(X, Y)|Y]$ and it's equal to $g(Y)$ where $g(y) = E[f(X, y)]$. Does it hold that $E[f(X, Y)|\mathscr{F}] = g(Y)$ as well?
My problem is that the proof of the formula relies on the fact that if $B\in\sigma(Y)$ then $B = A^{-1}(Y)$ where $A$ is a Borel set. I can't use that approach here.
Yes, this is true. You can find it, for instance, as (10.17) in Resnick, A Probability Path. (Resnick only considers bounded Borel functions $f$, but it's an easy step from there to general $f$ by truncating and using dominated convergence.)
The first step is to consider the case $f(x,y) = 1_A(x) 1_B(y)$ where $A,B$ are Borel sets. In this case, since $1_B(Y)$ is $\mathcal{F}$-measurable we have $$E[f(X,Y) \mid \mathcal{F}] = E[1_A(X) 1_B(Y)\mid \mathcal{F}] = 1_B(Y) E[1_A(X) \mid\mathcal{F}] = 1_B(Y) E[1_A(X)]$$ whereas $g(y) = 1_B(y) E[1_A(X)]$ and thus the right side equals $g(Y)$.
Then you use a $\pi$-$\lambda$ or monotone class argument to get $f = 1_C(x,y)$ where $C \subset \mathbb{R}^2$ is Borel. After that, it's the "standard mantra" through simple functions and so on.
