Here are some of my notes showing $f_{*}$ is left adjoint to $f^{-1}$.
If we consider the category of subsets $Sub(S)$ of a set $S$, which is the corresponding category of the partial ordered set $(\mathcal{P},\subseteq)$. A function $f: S \to T$ leads to two functors $f^{-1}: Sub(T) \to Sub(S)$ and $f_{*}:Sub(S) \to Sub(T)$. It turns out that $f_{*}$ is left adjoint to $f^{-1}$. Observing that $A \subseteq f^{-1}(f_{*}(A))$ for all $A$, we don't need to specify the unit of the adjunction. It suffices to check then that $A \subseteq f^{-1}(B) \iff f_{*}(A) \subseteq B$.
I have a doubt. Is it not sufficient to prove $A \subseteq f^{-1}(B) \implies f_{*}(A) \subseteq B$ if one knows that the unit and counit formulation of an adjunction are equivalent?
If we would prove $$ A \subseteq f^{-1}(B) \quad \Longleftrightarrow \quad f_*(A) \subseteq B, \tag{1} $$ then this would already prove that it is an adjunction of posets, since this correspondence is automatically natural. So that means there is no need to prove anything about a (co)unit.
Going for the unit approach, it would indeed also be enough to prove $A \subseteq f^{-1}(f_*(A))$ and $A \subseteq f^{-1}(B) \implies f_*(A) \subseteq B$. This works via the general machinery as you mentioned in the comment below your question. Alternatively, we can just prove $(1)$. The forward direction we have already. For the other direction, assume $f_*(A) \subseteq B$. Then $f^{-1}(f_*(A)) \subseteq f^{-1}(B)$, and so $A \subseteq f^{-1}(f_*(A)) \subseteq f^{-1}(B)$ as required.