How does the axiom of regularity make sense?
As an example, let $X = \{1,2\}$ (where $0=\varnothing, 1= \{0\}, 2=\{0,1\}$). Then $$X\cap 1 = \varnothing$$ $$X\cap 2 = \{1\}$$ Due to the first equality, our set satisfies the axiom of regularity. $X$ being disjoint with $1$ does not imply that $X$ does not contain $1$; $X\cap 1 = \varnothing$ is a completely different statement from $1\notin X$. Indeed, the former holds, as the only element of $1$ is $0$, which is not an element of $X$.
In a nutshell, the axiom forbid "loops".
Consider he case $A= \{ a_0, a_1, a_2 \}$.
If the axiom does not hold, we have the possibility that: for all $i$ : $a_i \cap A \ne \emptyset$.
1) Consider $a_0$; if the intersection of $a_0$ and $A$ is not empty, it must be an element of $A$.
Thus, three possibilities: i) $a_0 \in a_0$: loop; ii) $a_1 \in a_0$; iii) $a_2 \in a_0$.
Two possibilities left to avoid loops.
2) Consider $a_1$; again, if the intersection of $a_1$ and $A$ is not empty, we have: i) $a_0 \in a_1$; ii) $a_1 \in a_1$: loop; iii) $a_2 \in a_1$.
But $a_0 \in a_1$ and the previous $a_1 \in a_0$ form a loop: $a_0 \in a_1 \in a_0$.
Thus what remain is: $a_2 \in a_0$ and $a_2 \in a_1$.
Now: 3) Consider $a_2$; we have: i) $a_0 \in a_2$; ii) $a_1 \in a_2$; iii) $a_2 \in a_2$: loop.
Again, with: $a_0 \in a_2$ and $a_2 \in a_0$ we have a loop and the same with $a_1 \in a_2$ and $a_2 \in a_1$.
Your confusion seem to be due to levels of containment.
The members of a set is only those elements that are directly contained in the set. By this I mean that a set may contain other sets, but that does not make the elements of that set a member of the outer set.
This is a bit in contradiction in how we think of containment in ordinary life. Let's for example say you have a box of chocolates in your kitchen drawer. Then in ordinary life you would say that the drawer contains chocolates, but in mathematics terminology it doesn't it only contains a box (which in turn contains chocolates). Now you see that in mathematical terms the kitchen drawer and the box of chocolates are disjoint (but not in ordinary language).
To imagine a set that violates the axiom of regularity would probably need you to have some form of self containment. That is a container that in some way contains itself - which is quite contra intuitive, it doesn't fit well with the way we think of things.