On the meaning of the Axiom Of Regularity in terms of " repeatedly taking the union" . 
"whenever I try to chase down a chain of members, it must stop at some finite stage. You can think of it in this way. We have a string of sets $x_1,x_2,x_3...$ where each is a member of the preceding one; that is: $.....x_3 \in x_2 \in x_1$. This we shall call a descending membership chain. Then the axiom (Axiom of Foundation/Axiom of Regularity) is this: any descending membership chain is finite"

in Crossley et alii , What is mathematical logic? OUP, 1972 (Chapter 6 " Set Theory", pp. 62-63)

So the axiom is aimed at ruling out the possibility of having such a descending chain. 
Just before this passage, the author introduces the question by considering the fact that when I have a family of sets I can take the union of this family... that is I can chase the members of the members of this family ... 
Is the axiom equivalent to saying that we want the operation of taking the union of the union of the union ... of a family of sets to be stationary at some stage or that  this operation finally results in the empty set? Or neither perhaps? 

Not really.
If some set $x$ is $n$ levels "down" in $A$, then in $\bigcup A$ is is only $n-1$ levels down and hence in $\underbrace{\bigcup\cdots \bigcup}_nA$, it reaches the "surface". But this doesn't mean that everything reaches the "surface" together.
For example, let $A_{n,0}=n$, $A_{n,k+1}=\{A_{n,k}\}$, and $X=\{\,A_{n,n}\mid n\in\omega\,\}$.
Then we have
$$\underbrace{\bigcup\cdots\bigcup }_nX=\{\,A_{m,m-n}\mid m\ge n\}\cup (n-1)$$
so this sequence is not stationary (but of course $X$ is regular).
