Right-adjoint to the inverse image functor

Powersets, as posets / categories, are self-dual via taking complements. This implies that the right adjoint is the complement of the image of the complement.

A conceptual way to think about the left and right adjoints of taking inverse image is that they are given by fiberwise existential vs. universal quantification. That is, if $f : X \to Y$ is a map of sets and $f^{\ast} : 2^Y \to 2^X$ the inverse image map, its left adjoint takes a subset $S \subseteq X$ to the set

$$T = \{ y \in Y : \exists x \in f^{-1}(y) : x \in S \}$$

while its right adjoint takes a subset $S \subseteq X$ to the set

$$T = \{ y \in Y : \forall x \in f^{-1}(y) : x \in S \}.$$

Following your thoughts, we look for a functor $F:\mathcal P(X)\to\mathcal P(Y)$ which satisfies $$\hom_{\mathcal P(X)}(f^{-1}(V),\,U) \simeq \hom_{\mathcal P(Y)}(V,\,F(U))\,.$$ Since both categories in question are partial orders by inclusion, this means exactly that $$f^{-1}(V)\subseteq U\ \iff\ V\subseteq F(U)$$ for all $U\subseteq X,\ V\subseteq Y$.
This suggests $F(U):=\bigcup\{V\,:\,f^{-1}(V)\subseteq U\}\ =\ \{y\in Y\,:\,f^{-1}(y)\subseteq U\}$. $\ $(See also the comments.)

Note that, when viewing $f:X\to Y$ as an $Y$-indexed collection of its fibers $\{f^{-1}(y)\}_{y\in Y}$, then $F(U)$ just picks the indices of those fibers which are fully contained in $U$.