(This should fix an earlier attempt...) I don't see how one can prove that $V$ is the union of that particular chain, but here is a variant, from which we still get finite axiomatizability.

Suppose for a warm-up that the model $(M,j)$ of wholeness has wellfounded $\omega$. Define the critical sequence $\left<\kappa_n\bigm|n<\omega\right>$ externally (not in $M$). Claim: this sequence is cofinal in $\mathrm{OR}^M$. Otherwise let $\delta$ be an ordinal such that $\kappa_n<\delta$ for all $n<\omega$; then $k=j\upharpoonright V^M_{\delta+2}:V^M_{\delta+2}\to j(V^M_{\delta+2})$ is elementary and $k\in M$.
Since $M$ has wellfounded $\omega$, it follows
that $\sup_{n<\omega}\kappa_n\in M$,
and so we can take $\delta=\sup_{n<\omega}\kappa_n$, and then $j(\delta)=\delta$ and $k$ contradicts Kunen in $M$. So, since the sequence is cofinal in $\mathrm{OR}^M$, $M$ is the (external) union of the elementary chain $\left<V^M_{\kappa_n}\right>_{n<\omega}$, hence models ZF.

If $M$ has illfounded $\omega$, I don't see that we can make the same conclusion about the elementary chain (and see the comment above for a counterexample).
But I claim that there is an "internal critical sequence"
$\left<\kappa_n\bigm|n\in X\right>$ for some initial segment $X$ of $\omega^M$ (possibly a proper cut) such that each proper segment of the sequence is in $M$, and then this sequence is cofinal in $\mathrm{OR}^M$
and $M$ is the union of the corresponding elementary chain of $V^M_{\kappa_n}$, so we still get a version of the desired thing.
For given $\delta\in\mathrm{OR}^M$, we have $j\upharpoonright V^M_\delta\in M$, and can use this, working in $M$, to define some critical sequence $\left<\kappa^\delta_n\bigm|n\in X^\delta\right>$, with $X^\delta$ the largest possible initial segment of $\omega^M$. If $\delta<\beta$
then the sequence defined using $\delta$ is an initial segment of that defined with $\beta$. So let $X=\bigcup_{\delta\in\mathrm{OR}^M}X^\delta$, and $\kappa_n=\kappa^\beta_n$ for $n\in X$. Note that $X$ has no largest element. Now suppose that there is $\delta\in\mathrm{OR}^M$ such that $\kappa_n<\delta$ for all $n\in X$. Using $j\upharpoonright V^M_\delta$ we get that $X=\omega^M$ (using an overspill argument). But then we have $\delta'=\sup_{n\in\omega^M}\kappa_n\in M$ (computed from $j\upharpoonright V^M_\delta$), and can contradict Kunen in $M$.

Now observe that $M\models$"$V^M_{\kappa_m}\preccurlyeq V^M_{\kappa_n}$"
whenever $m<n$ with $m,n\in X$: This is just because $M\models$"$V^M_{\kappa_0}\preccurlyeq V^M_{\kappa_1}$", and so working in $M$, where we have $j\upharpoonright V_\delta$ for arbitratily large $\delta$s, by induction, we get $M\models$"$V^M_{\kappa_i}\preccurlyeq V^M_{\kappa_{i+1}}$" for each $i\in X$, which also by induction in $M$ is enough. It follows that (outside $M$) $V^M_{\kappa_m}\preccurlyeq V^M_{\kappa_n}$ for $m,n$ as above. Therefore $M$ is the union of this elementary chain.

Therefore (in either case) we get $M\models\mathrm{ZF}$, and hence finite axiomatizability of WA0.

Note this seems to leave the possibility that $X\subsetneq\omega^M$.
And in fact this can happen. Just start with a model $(M,j)$ of WA0
with $\omega^M$ illfounded and $\omega\subsetneq X$ where $X$ is as above (we get this by taking an ultraproduct of a countable model as in the comment above). We may assume $X=\omega^V$. Let $X'$ be any proper cut of $\omega^V$ and $V'=\bigcup_{n\in X'}V^V_{\kappa_n}$ where $\left<\kappa_n\bigm|n\in X\right>$ is as above. Let $j'=j\upharpoonright M'$. Then note that $j':M'\to M'$ is cofinal and $\Sigma_0$-elementary, since $M'$ is the union of the elementary chain $\left<V^M_{\kappa_n}\right>_{n\in X'}$. But then $M'\models\mathrm{ZF}$, so it follows that $j'$ is in fact fully elementary. And the other axioms of WA0 easily hold for $(M',j')$. Now if we start with $(M',j')$, we would get $X'$ via the process above, but $X'\subsetneq\omega^M=\omega^{M'}$.

Remark: It looks to me like essentially the same thing works for WA$n$.
E.g. for $n=1$, use WA$0$ + the axiom "For every set $x$ and $\Sigma_1$ formula $\varphi$ and parameter $p$, there is a set $y$ such that for all $z\in x$, we have $z\in y$ iff there is $\alpha\in\mathrm{OR}$ and a satisfaction relation for $(V_\alpha,j\upharpoonright V_\alpha)$ witnessing that $(V_\alpha,j\upharpoonright V_\alpha))\models\varphi(p,z)$". This incorporates possibly nonstandard formulas $\varphi$ in the usual way.