For question 1, let's forget about the poset entirely for the moment (it will come back later). Suppose we have a ground model $\mathcal{M}\models\mathsf{ZF}$, which for simplicity I'll assume is transitive. Now we can define the $\mathcal{M}$-class of names over $\mathcal{M}$ as the smallest $\mathcal{M}$-class $\mathcal{C}$ with the following properties:
$\emptyset\in\mathcal{C}$.
If $A\in\mathcal{M}$ and $A\subseteq\mathcal{C}$, and $f\in \mathcal{M}$ is any function with domain $A$, then $$\nu=\{\langle a, b\rangle: a\in A,b\in f(a)\}$$ is in $\mathcal{C}$.
Similarly, for every $X\subseteq\mathcal{M}$ (note that we're not assuming $X\in\mathcal{M}$ here!) and $\nu\in\mathcal{C}$ we can define $\nu[X]$ recursively as $$\nu[X]=\{\mu[X]: \exists b\in X(\langle\mu,b\rangle\in \nu)\}.$$ The point is this:
Proposition: If $\mathcal{N}$ is a transitive model of $\mathsf{ZF}$ with $X\in \mathcal{N}$ and $\mathcal{M}\subseteq\mathcal{N}$, then for each $\nu\in\mathcal{C}$ we have $\nu[X]\in\mathcal{N}$.
Basically, everything "built by" a name has to be in the resulting model. The proof of the above fact is just a quick application of transfinite recursion, and this application reinforces the whole "names-as-recipes" idea in the linked answer.
This then raises a natural idea, letting $\mathcal{M}$ be as above:
Guess: There is some notion of "reasonable" such that for reasonable sets $X$ (and perhaps further assumptions about $\mathcal{M}$, e.g. countability) we have $\{\nu[X]:\nu\in\mathcal{C}\}\models\mathsf{ZF}$ (and so in particular is the smallest transitive model of $\mathsf{ZF}$ containing $\mathcal{M}$ as a submodel and $X$ as an element).
This is where the whole "generic filter through some poset $\in\mathcal{M}$" finally appears; it's the reasonableness-criterion needed for making the guess above work. But right up until this final moment, we don't have to think about posets/filters at all.
Of course, the above assumes that the "names-as-recipes" idea actually clicks. For this I think the best thing to do is actually write out some translations. For instance:
Fix some set $u\in\mathcal{M}$. Show that, for each $a\in\mathcal{M}$, there is a $\nu\in\mathcal{C}$ such that for every $X\subseteq\mathcal{M}$ with $u\in X$ we have $\nu[X]=a$. (This $u$ is basically a "default-yes" element; its role is taken by $\mathbb{1}$ in the poset.)
Again assuming that we're restricting to $X\ni u$ for some $u\in\mathcal{M}$ as above, write out the name-over-$\mathcal{M}$ corresponding to the recipe (from the linked answer) $$=\{\{\{...\}\} \mbox{($n$ many brackets)}: n\in X\}.$$
(At this point you might be wondering why I didn't use recipe-rules where the right-hand-component is/isn't a subset of, rather than an element of, the set $X$ involved; the answer is just that I wanted to stick closer to the usual definition of names.)
- OK, now assuming that $X$ is a filter through some poset $\mathbb{P}\in\mathcal{M}$ with greatest element $\mathbb{1}$, given $\nu,\mu\in\mathcal{C}$ find a $\rho\in\mathcal{C}$ such that we are guaranteed to have $\rho[X]=\mu[X]\cap\nu[X]$. We're still not assuming genericity here, but we are going to see how the poset structure helps us with our name constructions.
