Let $f: X \to Y$ be a function. Let $\mathcal{F}$ be a filter on $Y$ and for all $F \in \mathcal{F}: F \cap f(X) \neq \emptyset$. Then, $\mathcal{F} \subseteq \operatorname{stack} [f(f^{-1}(\mathcal{F})]$ and if $f$ is surjective, then $\mathcal{F} = f(f^{-1}(\mathcal{F})$
My attempt:
Let $F \in \mathcal{F}$, then $F \supseteq f(f^{-1}(F)) \in f(f^{-1}(\mathcal{F}))$. Hence, $F \in \operatorname{stack}[f(f^{-1} (\mathcal{F}))]$. This shows that the inclusion is true.
If $f$ is surjective, then it is trivial that $\mathcal{F} = f(f^{-1}(\mathcal{F}))$, since for any $F$, $f(f^{-1}(F)) = F$.
Where did I go wrong? I never used that $\mathcal{F}$ was a filter, or that $F \cap f(X) \neq \emptyset$ (the last condition is irrelevant if $f$ is surjective though)
The condition is used to ensure that for a filter $\mathcal{F}$ on $Y$ the set $f^{-1}[\mathcal{F}] = \{f^{-1}[F] : F \in \mathcal{F}\}$ is actually a filter on $X$. It's used for the non-triviality axiom $\emptyset \notin f^{-1}[\mathcal{F}]$.
For a filter $\mathcal{B}$ on $X$ we define $f[\mathcal{B}] = \{f[B]: B \in \mathcal{B}\}$, and this is a filterbase on $Y$, so that $\operatorname{stack}(f[\mathcal{B}])$ is a filter on $Y$.
Indeed, the first inclusion is trivial: if $F \in \mathcal{F}$, then $f[f^{-1}[F]] \in f[f^{-1}[\mathcal{F}]]$ and as $f[f^{-1}[F]] \subseteq F$ by definition, indeed $F \in \operatorname{stack}(f[f^{-1}[\mathcal{F}]])$, and so $\mathcal{F} \subseteq \operatorname{stack}(f[f^{-1}[\mathcal{F}]])$. No more to it than that.
And if $F$ is surjective, we always have $f[f^{-1}[F]] = F$ which shows the reverse inclusion as filters are closed under supersets: $A \in \operatorname{stack}(f[f^{-1}[\mathcal{F}]])$ means there is some $F \in \mathcal{F}$ with $f[f^{-1}[F]] \subseteq A$ so $F \subseteq A$ and so $A \in \mathcal{F}$. Hence $\mathcal{F} \supseteq \operatorname{stack}(f[f^{-1}[\mathcal{F}]])$ and equality. We indeed use little of the filter axioms.