The union in your quote should probably be
$$ \omega_0 \cup \omega_{\omega_0} \cup \omega_{\omega_{\omega_0}} \cup \cdots $$
is the smallest solution to the equation $\alpha=\omega_\alpha$, which is also mentioned in your quote.
I don't think that particular set has any nicer description than the two you already have, so asking "what it is" cannot really produce any answer other than the definitions you already have. It's a certain initial ordinal, with the properties you see before you.
ZFC can prove that a set with these properties exists, by defining a function from naturals to ordinals as
$$ F(0) = \omega \\ F(n+1) = \omega_{F(n)} $$
Then applying the Axiom of Replacement to $\omega$ and this function gives you
$$ \{ \omega_0, \omega_{\omega_0}, \omega_{\omega_{\omega_0}}, \ldots \} $$
and the Axiom of Unions applied to this gives
$$ \alpha = \omega_0 \cup \omega_{\omega_0} \cup \omega_{\omega_{\omega_0}} \cup \cdots $$
To see that this set satisfies $\alpha = \omega_\alpha$, note that it is clearly a limit ordinal, and for limit ordinals $\omega_\alpha$ is defined as the smallest ordinal that is $\ge$ $\omega_\beta$ for all $\beta<\alpha$ -- which by well-known properties of ordinals is the union of all these ordinals.
Since $\alpha$ is the union of a sequence of ordinals, and $\omega_-$ of each of those ordinals is also in the sequence, this means that $\alpha$ and $\omega_\alpha$ are both the union of essentially the same increasing sequence of ordinals; therefore they have to be equal.