Theorem 11.10 of Double helix in large large cardinals and iteration of elementary embeddings by Kentaro Sato proves that if $\kappa$ is the critical point of an I2 elementary embedding $j: V \to M$ and $U$ is the ultrafilter generated by $j$, then there is a set $T \in U$ such that every subset of $T$ that is of order type $\omega$ is the critical sequence of an I3 embedding. The proof builds such embeddings from approximations $f_\sigma: \mathcal{M}_{\sigma(n-2)} \to \mathcal{M}_{\sigma(n-1)}$, where $\sigma: n \to T$ is an increasing function, and for the purpose of this question we can take $\mathcal{M}_{\sigma(i)}$ to be $V_{\sigma(i)+1}$ (by $f_\sigma^{(i)}$, the paper means $f_\sigma^i (f_\sigma \upharpoonleft \mathcal{M}_{\sigma(n-i)})$, and similarly for $e^{(i)}$ but I consider this notation unnecessary). I don't understand the following step of the proof:
For $g \in ^{\omega \uparrow}T$, by $g(N) \in T \subset S^{(n)}$, we have $g(n) \in S_{g \upharpoonright n}$ for all $n \in \omega$. Define $e^{(n)} = \bigcup_{i \gt n} f_{g \upharpoonright n}^{(n)} : \bigcup_{i \ge n-1} \vert \mathcal{M}_{g(i)} \vert \to \bigcup_{i \ge n} \vert \mathcal{M}_{g(i)} \vert$. Then, by definition, $\vec{e}$ is an iteration sequence through $\langle \mathcal{M}_{g(n)} \vert n \lt \omega \rangle$
It seems that we only need the approximations $f_\sigma$ for this step, without proving that $M$ sees I3 embeddings approximating $j$. Instead, we could define a tree of elementary embeddings $f_\sigma: \mathcal{M}_{\sigma(n-2)} \to \mathcal{M}_{\sigma(n-1)}$ such that $f_\sigma$ extend to $n$-huge embeddings for every $n$. By the dependent choice principle, that tree should have an infinite branch. However, if this simplification worked, it would prove that every I3 critical point is a limit of I3 critical points, which would mean that I3 (and thus I2) would would be incontistent. Why does this simplification fail?