Stratification of Bisimilarity : $\sim$ does not coincide with $\sim_\omega$
I found the answer of my question when writting it, which once again prove the effectiveness of writting down properly the question we have.

To show that $s \sim_1 t$, to map $a^\omega$, one can take any $a^i$ we want (e.g. $a^0$): by definition,
$\sim_0 \overset{\text{def}}{=} W\times W$, and therefore we indeed have that $s\overset{a}{\rightarrow} a^0$ and $a^0\sim_0 a^\omega$.
Something to notice is that, for each $n$, we have $a^\omega \sim_n a^m$ for $m\geq n$. Therefore, to show that $s\sim_{n+1} t$, to show item (2), one can simply take $a^n$ as $s^\prime$ when we take $a^\omega$ as $t^\prime$. The rest of the proof is trivial.