I agree with Anindya—the definition of \$$f_!:PX\to PY\$$ that makes it the right adjoint to \$$f^\ast: PY\to PX\$$ should be \$f_!(S) = \\{y\in Y : \text{ for all }x\in X, \text{ if }f(x)=y\text{ then }x\in S\\}.\$

One way to see how the dual quantifiers "for all" and "there exists" arise is to derive \$$f_!\$$ from \$$f_\ast\$$ using complements: first note that \$$f^\ast\$$ commutes with complements in the sense that \$$f^\ast(Y\setminus T) = X\setminus f^\ast(T)\$$. Then we find that \$f^\ast(T)\subseteq S \iff X\setminus S \subseteq f^\ast(Y\setminus T)\$
(by taking the complement of both sides, which reverses the inclusion)
\$\iff f_\ast(X\setminus S)\subseteq Y\setminus T\$
(using the adjunction between \$$f_\ast\$$ and \$$f^\ast\$$)
\$\iff T\subseteq Y\setminus f_\ast(X\setminus S)\$
(by taking the complement of both sides again). Therefore the operation \$$f_!(S) := Y \setminus f_\ast(X\setminus S)\$$ is a right adjoint for \$$f^\ast\$$.

This double-complement expression tells you that if \$$f_\ast\$$ is described by a quantifier (in our case \$$\exists\$$) then \$$f_!\$$ should be described by the dual quantifier (in our case, \$$\neg\exists\neg = \forall\$$). Indeed, we can check carefully that one interpretation of \$$f_!(S)\$$ is the set of all \$$y\$$ for which it's *not* true that some \$$x\$$ *not* in \$$S\$$ maps to \$$y\$$.