Quasi-compactness is local on the target and survives base change

By F1 and F2, quasi-compactness of $f$ implies the condition for every affine open, which implies the condition for any chosen affine cover. Suppose conversely that $S=\bigcup U_i$ is such a cover. For an arbitrary affine open $V\subset S$, principal opens in the $U_i$ contained in $V$ cover $V$. By F2 choose finitely many of them, say $W_1,\dots ,W_n$.

For each $W_j=D_{U_i}(a)$ take a finite affine cover of the quasi-compact $f^{-1}(U_i)$. Its intersection with $f^{-1}(W_j)$ is principal in each affine chart, being the nonvanishing locus of the image of $a$, and is affine. Hence $f^{-1}(W_j)$, and then $f^{-1}(V)$, is a finite union of affines, thus quasi-compact by F2. Every quasi-compact open of $S$ has a finite affine cover; its inverse image is consequently quasi-compact. This is exactly F1.

For $S^{\mathrm{\prime }}\to S$, cover $S^{\mathrm{\prime }}$ by affines $V^{\mathrm{\prime }}$ mapping into affines $U\subset S$. A finite affine cover of $f^{-1}(U)$ pulls back by F3 to a finite affine cover of the inverse image of $V^{\mathrm{\prime }}$. It is quasi-compact by F2. The criterion just proved gives quasi-compactness of the base-changed morphism. Empty covers, zero coordinate rings, and singleton covers are included. Composition of quasi-compact morphisms also follows directly from F1: pull back a quasi-compact open first by the second morphism and then by the first; both successive inverse images are quasi-compact.