No, this has nothing to do with regular languages (and likewise nothing to do with regular cardinals).
As for intuition, I find it much easier to think about what it would mean for a set not to be regular:
A set $x$ is non-regular if there is an $y$ which contains $x$ as an element, such that for every $w\in y$ there is a $z\in w$ that is also in $y$.
This means, in particular, that we can take $w$ to be $x$, and find a chain of elements
$$ x \ni w_1 \ni w_2 \ni w_3 \ni \cdots $$
where at each step $w_i$ can be chosen to be in $y$, which allows us to continue the chain indefinitely.
So if a set $x$ is not regular, there is a way to construct an infinitely descending $\in$-chain starting from $x$. (This can be carried out inside the set theory itself if we have the Axiom of Choice, and at the metalevel if we have choice there).
Conversely, if $x$ is regular, then there cannot be any infinite chain
$$ x \ni x_1 \ni x_2 \ni x_3 \ni \cdots $$
-- or at least such a chain cannot be known inside set theory in the form of a function defined on $\omega$ that maps each $i$ to $x_i$ -- because the range of that function would work as an $y$ that certifies $x$ as non-regular.
It is still possible that an element of a model of set theory can be regular, yet there is an infinite chain that we can see at the metalevel, that is, looking at the model from outside. (This will be the case for any model of set theory that contains non-standard integers, for example). But in that case the chain cannot be encoded in toto as an object within the model.