Let \(f : X \to Y\) be separated, let \(Z\) be a reduced scheme over \(Y\), and let \(g, h : Z \to X\) be \(Y\)-morphisms agreeing on a dense open \(U \subseteq Z\). Then \(g = h\).
This is the good property to name, and the proof shows why both hypotheses are there. The pair \((g,h)\) gives \(Z \to \fiberprod{X}{Y}{X}\), and the locus where \(g\) and \(h\) agree is the preimage of the diagonal. Separatedness makes that preimage closed; it contains the dense \(U\), so it is all of \(Z\) topologically. Reducedness upgrades the topological statement to an equality of morphisms, because a closed subscheme of a reduced scheme with the whole space as support is the whole scheme.
So the analogue for quasi-separated is weaker in exactly the way the diagonal is: quasicompactness of the diagonal gives a retrocompact agreement locus, not a closed one, and the conclusion fails. Quasi-separatedness is what one assumes to make cohomology and direct images behave, not to make morphisms equal.