I'm following Jech and in page 518 there is the following corollary (26.11):
let $G$ be generic for Levy collapse $P$ and $X$ countable set of Ordinals in $M[G]$. Then there exists a $V[X]$-generic $H$ on $P$ such that $V[X][H]=V[G]$.
Following the proof they used corollary 26.10 (paraphrasing to get only the relevant part):
Let $G$ be generic on $Col(\omega,\lambda)$ and $X$ set of $M[G]$ Ordinals, if $M[X]$ thinks that $\lambda$ is uncountable, then there is $M[X]$ generic on $Col(\omega,\lambda)$, $H$, such that $M[X][H]=M[G]$
In which they immediately stated "$V[G]$ is a generic extension of $V[X]$ by $Col(\omega, |\lambda|^{V[X]})^{V[X]}$"
I understand why we can get a generic $H$ over $V[X]$ by $Col(\omega, |\lambda|^{V[X]})^{V[X]}$, but I am lost in how to get equality.
(Side note: I'm much more comfortable using poset forcing, and not Boolean algebra forcing like Jech does it, so I would appreciate it if the answer uses posets)
The quotient forcing $Q$ in $V[X]$ that gets you to $V[G]$ has size at most $\kappa = \vert \lambda \vert^{V[X]}$, since it is literally a subset of $\operatorname{Coll}(\omega, \lambda)$. But if $\lambda$ is still uncountable in $V[X]$, this means that $Q$ will collapse $\kappa$ to $\omega$ and by Lemma 26.7, $Q$ is isomorphic to $\operatorname{Coll}(\omega, \kappa)$.
The same argument, namely that $Q \subseteq \operatorname{Coll}(\omega, \lambda)$ is then also used to show that $Q$ is countable when $\vert \lambda \vert = \omega$.