Distributed MST Construction in O(log log n) Rounds in a Clique
I think I've found an answer, so will post here what I have. My proof seem to cover both scenarios I've mentioned. I still think their proof is incomplete. But in case I'm wrong, I would love to be corrected.

Similar to them assuming there is $e'$ outgoing from $F_1$, I assumed there was $e'$ outgoing from $F_3$ that is in $\hat{F_{1}}$, of lower weight than $e$.

If it was chosen, than it happened before choosing $e$, since its lighter. But then we get a contradiction to $e'$ being outgoing of $H_1$. So it wasn't chosen. If it wasn't chosen because it created a cycle, then it's yet again a contradiction to it being outgoing from $H_1$, since if it created a cycle, then $F_3, F_4$ were already merged by the time we reached $e'$, and therefore when we reached $e$ as well.

Therefore we're only left with the option that it was deemed "dangerous" (and therefore removed). Then either $e'\notin\mathcal{A}\left(F_{3}\right)$ or $\hat{F_{3}}$ was finished by the time we reached $e'$ .

If $\hat{F_{3}}$ was finished , then by the time we get to $e$ (which happens after $e'$, as $e'$ is lighter). But we assumed $F_3$ is part of $H_1$ by the time we reach $e$, therefore $\hat{F_{1}}$ is also finished, which is a contradiction to it being unfinished.

Then $e'\notin\mathcal{A}\left(F_{3}\right)$. This can happen for two reasons (as per their proof):


There exists a lighter (than $e'$) $e''$ in $\mathcal{A}\left(F_{3}\right)$ (such that $e''$ touches the same base fragments $e'$ touchs, which means its basically the same logical edge, just of lower weight). Therefore since $\hat{F_{3}}$ wasn't finished, and there were no cycles, $e''$ was added earlier than $e'$ and therefore than $e$. But then by the time we reach $e$, $e'$ is no longer outgoing from $H_1$. Contradiction yet again.
Then there exists $\mu$ lighter edges in $\mathcal{A}\left(F_{3}\right)$. Let $e''$ be the heaviest there (that is still lighter than $e'$ and therefore than $e$). Therefore after inspecting $e''$, $\hat{F_{3}}$ is deemed finished, which is yet again a contradiction.