We will always work with finite-type, smooth schemes over a field $k$. Let $\pi: Y \to X$ be an etale map of $k$-schemes. Let $Z$ be another $k$-scheme admitting a morphism $f: Y \to Z$. Suppose that there exists a scheme $Y'$ and an etale map $\pi': Y' \to Y$ such that $\pi' \circ f : Y' \to Z$ descends to a map $X \to Z$. Is it necessary that $f$ also descends to a morphism $X \to Z$? If not, can one give me a counterexample?
If I understand correctly, you have a morphism $g:X\rightarrow Z$ such that $g\circ \pi \circ \pi '=f\circ \pi '$, and you ask whether this implies $g\circ \pi =f$. You have to assume $Y$ connected (or at least that $\pi '(Y')$ hits every component of $Y$). Then since an étale map is open, $g\circ \pi $ and $f$ coincide on an dense open subset of $Y$, hence they are equal.
Calling this descent is a bit pompous, by the way...