In The Principles of Mathematical Analysis, 3rd Ed. by Rudin, while proving Theorem 4.14 (if $f$ is a continuous function from a compact metric space $X$ to a metric space $Y$, then $f(X)$ is compact), he mentions in passing that $f\circ f^{-1}(E) \subset E$, where $E \subset Y$, and that the equality need not hold, as in, it isn't necessary that $f\circ f^{-1}(E) = E$.
Is this only because $E$ being any subset of $Y$ may include points that aren't in the range of $f$? Otherwise, the equality always holds, right?
Yes, $f^{-1}(E)=\{x\in X; f(x)\in E\}$, so if you take a point $a\in E\setminus Im(f)$ (if exists), then $a\notin f\circ f^{-1}(E)$ as you said.