The premise of the question I ran across here is the fact that if $M$ is some model of $\mathsf{ZFC}$ (whose reals I denote as $\mathcal{P}(\omega)^{M}$) then there always exists some “bigger” model of $\mathsf{ZFC}$ $M’$ in which $\mathcal{P}(\omega)^{M}$ is actually countable
But I wonder about clarifying “countability” in the “bigger” model $M’$: Specifically, if $M$ has standard $\omega$ (in the sense discussed here), does that mean that $M’$ is (or can be) also $\omega$-standard?
This is slightly misleading. The question talks about the Levy collapse. Namely, given a model of $\sf ZFC$, $M$, and any set $x\in M$, there is a forcing notion $\Bbb P\in M$ such that any $M$-generic filter for $\Bbb P$ will introduce a bijection between $x$ and $\omega$.
Being a generic extension, it will not add ordinals, in particular, it will not add natural numbers.
Of course, the question is why is there such generic filter. And the answer is that we can only guarantee the existence of a generic filter when we assume $M$ is countable, or if we have some additional assumptions in the universe (e.g. some forcing axioms).
Still, once you are comfortable enough with forcing, it becomes a hindrance to always talk about a transitive model of enough axioms of $\sf ZFC$, and instead we move to forcing "over the universe", as the machinery is internal anyway.
One can make the argument that it is possible to take Boolean-valued ultrapowers of an arbitrary model and make those into new $2$-valued models, which are simply not well-founded anymore. But then, especially in the case we are using a forcing notion that is collapsing things to be countable, we will have little to no control over whether or not we have added integers to the new model, but it will certainly not be a well-founded and transitive model.
