Let $N \prec \left(L_{\omega_1}, \in\right)$ be a countable elementary submodel.
In Jech Set Theory he states:
$N$ is transitive. Let $X\in N$. Let $f$ be the $<$-least mapping of $\omega$ onto $X$. Since $f$ is definable in $(L_{\omega_1}, \in)$ from $X$, $f$ is in $N$. Hence $f(n)\in N$ for all $n<\omega$ and we get $X\subset N$.
What I don't understand is the way $f$ is chosen and the reasoning behind why it is definable and how.
Can you apply the above to an elementary submodel $N\prec L_{\omega_2}$ s.t $|N|=\aleph_0$?
One of the main points of the argument is that every object in $L_{\omega_1}$ is countable internally. (Of course, $\omega_1=\omega_1^L$ in this problem.) That is, $L_{\omega_1}$ catches every set in itself is countable. Since $N\prec L_{\omega_1}$, $N$ also thinks every object is countable.
Thus this argument does not work if we replace $\omega_1$ with other ordinals, like $\omega_2$: every countable elementary submodel of $L_{\omega_2}$ would think there is an uncountable set, although it could not be a genuine uncountable set.
I explained why $f$ is definable in the comment, but let me summarize the point: $L_{\omega_1}$ thinks there is a definable global well-order (namely $<_L$.) Thus $N$ also thinks there is a global well-order which is definable. Hence the formula
$$\phi(f) :\equiv [f\text{ is onto from $\omega$ to $X$ and $f$ is the $<_L$-least function among them}]$$ is a first-order formula which defines an object (which is a function we desire.)