For a function $f: X \to Y$, if $Y-V$ is finite, when is $X - f^{-1}(V)$ finite?

As pointed out by @Cronus in the comments, it is sufficient that $f$ has the property that the preimage of every point is a finite set. I believe it is also a necessary condition.

Let's just consider surjective functions, just for convenience (it really doesn't change anything since any function $f:X \to Y$ can be 'made surjective' by replacing its codomain with its image.)

To see that the condition is sufficient, note that if some $f: X \to Y$ has the property, then for finite $Y \setminus V$, $$ X \setminus f^{-1}(V) = f^{-1}(Y \setminus V) = \bigcup_{y \in Y \setminus V} f^{-1}(y).$$ In other words, $X \setminus f^{-1}(V)$ is the union of a finite number of finite sets.

I believe this condition is also necessary, and here is a proof that seems to be correct. Assume that $f$ doesn't have the property, and thus there exists some $y$ with $f^{-1}(y)$ not finite. Taking $V = Y \setminus \{ y \}$, one then concludes that $$ X \setminus f^{-1}(V) = f^{-1}(y),$$ which is not finite. Importantly, we have that $f^{-1}(V)$ does not intersect $f^{-1}(y)$ since no $x \in X$ can be mapped both into $V$ and into $\{ y \}$. Thus by contradiction we have that $f$ must have the finite-to-one property.