Let $f: R \to S$ be a morphism of noncommutative rings. Let $$f_!:= S \otimes_R (-) : R \text{Mod} \to S \text{Mod}$$ denote the extension of scalars functor, and $$f^*: S \text{Mod} \to R \text{Mod}$$ the restriction of scalars functor. These form an adjunction $f_! \dashv f^*$.
Using the natural isomorphism $f^* \cong f^*S \otimes_S (-)$, I get $$f_!f^*W \cong S \otimes_R (f^*S \otimes_S W) \cong (S \otimes_R f^*S) \otimes_S W,$$ The $S$ on the left is an $(S,R)$-bimodule, and $f^*S$ is an $(R,S)$-bimodule, and it would be nice if we had $S \otimes_R f^* S \cong S$ naturally as $(S,S)$-bimodules, but I get the feeling this is not correct, since then we would have $f_! f^* W \cong W$ as $S$-modules, which seems to be wrong.
Can this be "simplified" further, and what would be the correct way of writing down and understanding the counit $$\varepsilon: f_! f^* \Rightarrow \text{id}_{S \text{Mod}}$$ of the adjunction here?
First of all, we do not even have an isomorphism between $S\otimes_R f^*S = S\otimes_R S$ and $S$ as left $S$-modules (take for example $S$ to be a field and $R$ to be a subfield to see this).
The counit is fortunately very easy to describe. We have the map from $S\otimes_R S$ to $S$ that sends $x\otimes y$ to $xy$, and as you noted, we can identify $f_!f^*W$ with $(S\otimes_R S)\otimes_S W$, which we can map to $W$ by sending $x\otimes y\otimes w$ to $(xy).w$.
That this map is in fact natural I leave as an (easy) exercise.