Define hierarchy as a family of sets well ordered by inclusion such that each successor is the powerset of its immediate predecessor, and the limit sets are the unions of all their predecessors.
A subhierarchy is a subfamily of a hierarchy
Now the elements of a subhierarchy are called as its ranks.
Now replacement seems to correspond to saying that every subhierarchy having the inclusion relation on it being isomorphic to a well ordering that is an element of one of its ranks, then that subhierarchy is a set.
I like to paraphrase the above paragraph as: any subhierarchy whose height comes from below, is a set.
if we add that principle to $\sf Z + \forall x \exists \alpha: x \in V_\alpha$, then would this prove Replacement?
Yes, that is enough.
The way we prove Replacement from Reflection is to take $\varphi$ which defines a function, given any $x$, reflect $\varphi$ to some $V_\alpha$ for a large enough $\alpha$, and then argue that the image of $x$ is a subset of $V_\alpha$, and therefore an element of $V_{\alpha+1}$.
The principle you describe allows us something similar. If $\varphi(x,y)$ defines a function, and if $A$ is any set, then the collection of the ranks of $\{y\mid\exists x\in A\,\varphi(x,y)\}$ must be bounded, since it is isomorphic to a well-ordering of a quotient of $A$. But that means that the collection itself is a set by means of Separation axioms to the appropriate $V_\alpha$.