If there is countable transitive model of ZFC, this model cannot capture all ordinals of ZFC. But we use it for stuffs like forcing.
But this seems to violate ZFC's axioms - for example, power set axiom (take, $\omega = \aleph_0$ and apply power set operator.).
So how can we use countable transitive model of ZFC, then?
On
The set that serves within the model as the power set of a set would not contain all subsets of the set, but only all subsets that belong to the model.
The theorem that the power set of an infinite set is uncountable would be true within the model simply because such a set would not be "internally countable", i.e. no enumeration of the set would be a member of the model.
On
Let $M$ be a countable transitive model of $\mathsf{ZFC}$. Obviously $\wp(\omega)^M$, the set that $M$ ‘thinks’ is the power set of $\omega$, is countable in $V$, since it’s a subset of $M$, but there is no $f\in M$ such that $M\models\text{'}f\text{ is a bijection between }\omega\text{ and }\wp(\omega)^M\text{'}$. Thus, from $M$’s point of view $\wp(\omega)^M$ is uncountable.
It’s hard at first, but you have to keep straight what $V$ ‘thinks’ and what $M$ ‘thinks’.
You are mixing internal and external point of views.
If $M$ is a countable transitive model of ZFC then it doesn't know that it is countable, and it certainly don't know that all its members are countable. That is to say, there are many sets $x\in M$ such that $M\models\aleph_0<|x|$. So while that in the full universe there is a bijection between $x$ and $\omega$ it is not an element of $M$.
When we use c.t.m for forcing we work internally when we define the forcing poset, and we argue externally when we prove that there is a generic set, and that the construction results in another countable transitive model of ZFC.
Somewhat related: