A common approach to forcing is to use countable transitive model $M \in V$ with $\mathbb{P} \in M$ and take a $G \in M$ (which always exists) to form a countable transitive model $M[G]$. Another approach takes $M$ to be countable such that $M \prec H_\theta$ for sufficiently large $\theta$ (and hence may not be transitive). For example, a definition of proper forcing considers such models.

Forcing with transitive models are quite convenient since many absoluteness results can be used to transfer properties of $x \in M[G]$ which hold in $M[G]$ up to $V$. If $M \prec H_\theta$ is not transitive, then it is not clear what type of property that $M[G]$ can prove about $x$ transfer to $V$. For instance, if $M[G] \models x \in {}^\omega\omega$, is $x \in {}^\omega\omega$ in $V$? Of course, one remedy could be to Mostowski collapse everything and then use the familiar absoluteness for transitive models. For $x \in {}^\omega\omega$, one could use the fact that $M \prec H_\theta$ implies $\omega \subseteq M$ and hence the Mostowski collapse of $M[G]$ would maps each real to itself and then use absoluteness to prove that $V \models x \in {}^\omega\omega$ as well. Is there a more direct way to prove these type of result rather than collapsing the forcing extension, which seem to suggest one should have started by collapse $M$ before starting the forcing construction.

So my questions are

1 First, if one chooses to work with countable $M \prec H_\theta$ are there any changes that need to made to the forcing construction and the forcing theorem as they appear in Kunen or Jech? Of course, the definition of a generic filter should be changed to meeting those dense sets that appear in $M$.

2 I am aware that if $G$ has master conditions, then $M[G] \prec H_\theta[G]$? Is $H_\theta[G]$ just the forcing construction applied to $H_\theta$? As $G$ is not necessarily generic over $H_\theta$, it is not clear to me that the forcing theorem need to apply to $H_\theta[G]$ (or a priori $H_\Theta[G]$ models any particular amount of $\text{ZF}- \text{P}$, but since $M[G] \prec H_\theta[G]$, actually $H_\Theta[G]$ would model as much as $M[G]$.) In general without addition assumption like master conditions, does the relation $M[G] \prec H_\Theta[G]$ still hold.

Also perhaps I am misunderstanding something, but since $\mathbb{P} \in M$, it appears that if $\theta$ is large enough, every $G \subseteq \mathbb{P}$ which is $\mathbb{P}$-generic over $M$ is already in $H_\Theta$. Would this not imply that $H_\theta[G] = H_\theta$ and hence $M[G] \prec H_\Theta$. Since $M \prec H_\theta$, $M$ and $M[G]$ models the exact same sentences. This surely can not happen.

Thanks for any help and clarification that can be provided.