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\\).