Suppose we force with a partial order $\mathbb{P}$ which lives in a model of set theory $M$. It's not too hard to show that if $M$ is countable, then there always is a filter $G$ in $V$ which is $\mathbb{P}$-generic over $M$. However, without this assumption there won't always be one. If $M$ contains everything in $(2^{\omega})^V$, for example, there is no generic filter for Cohen forcing.
So, how do we get around this. All the forcing arguments I see being with something like "Suppose $\mathbb{P}$ is a notion of forcing, and that $G$ is $\mathbb{P}$-generic over $M$." Is this an extra assumption, though? How do consistency arguments go through if not?
As mentioned in the comments, if you only care about consistency arguments, then there's no reason not to use a countable model where there are always generic filters available, although there are some other steps in getting from a forcing argument to a consistency proof that are more subtle.
But actually, consistency arguments don't really even need models at all. You can define the forcing poset $(P,\le, 1)$ and the forcing relation (really a relation schema) in $\sf ZFC$ and show that $\mathsf{ZFC}\vdash 1\Vdash_P\varphi$ whenever $\mathsf{ZFC}\vdash\varphi,$ and also that $\mathsf{ZFC}\vdash 1\not\Vdash_P\psi$ where $\psi$ is the continuum hypothesis, or whatever you're trying to prove independent.
(This can also be done with Boolean-valued models: we just replace $1\Vdash_P\varphi$ with $\lVert\varphi\rVert_B=1.$)
For reasoning about the "generic filter", we certainly don't have something we can define and prove is a generic filter over $P$, but what we do have is a name $\dot G:=\{(\check p,p): p\in P\}$ we can define such that $\mathsf{ZFC}\vdash \left[\mbox{ If $\mathcal{D}\subseteq \mathcal{P}(P)$ is a collection of dense sets, then $1\Vdash_P \forall D\in \check{\mathcal{D}}\;D\cap \dot G \ne \emptyset$ }\right]$ and $\mathsf{ZFC}\vdash \left[1\Vdash\mbox{$\dot G$ is a filter on $\check P$}\right].$ This allows us to carry out the genericity arguments we would normally need the generic filter for, and we can likewise cast statements like "c.c.c. forcings preserve regular cardinals" purely in terms of the forcing relation and prove that they hold in an appropriate sense.
The way the syntactic approach shakes out suggests that even when we're working with models, we should be entitled - in some sense - to pretend we have a generic filter/extension even when we don't literally. For a sense of just how far we can take this, see the first half of this nice paper by Hamkins and Seabold.
The upshot is that for an arbitrary (not necessarily standard/well-founded) model $(V,\in_V)\models \sf ZFC,$ with a forcing notion $P\in V,$ there is
such that:
So, modulo an elementary extension, generic filters and extensions do exist for arbitrary models. This generalizes both the usual forcing extensions of transitive models, as well as the more elusive forcing extensions of non-well-founded models: regardless of the well-foundedness of $V$, the above formalism works with $j$ as the identity if and only if $G$ is really a $V$-generic filter on $P$.
So there are many - probably too many - reasons why it's okay to just begin with something like "Suppose $\mathbb{P}$ is a notion of forcing, and that $G$ is $\mathbb{P}$-generic over $M$," as is typically done. Particularly, see "the naturalist account of forcing" section in the above paper.