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Let's begin by correcting, as Victoria does, your definition of what it means for $G\subset\mathbb{P}$ to be $M$-generic, in the case where $M\prec H_\theta$ is a possibly non-transitive elementary submodel of some $H_\theta$. You said to be generic means to meet every dense subset $D\subset \mathbb{P}$ with $D\in M$, but this is not the right definition. You want to say instead that $G$ meets every such dense set $D$ inside $M$. The point is that it could be that $G\cap D\neq\emptyset$, yet because it isn't transitive, $M$ can't see this.

Let's begin by correcting, as Victoria does, your definition of what it means for $G\subset\mathbb{P}$ to be $M$-generic, in the case where $M\prec H_\theta$ is a possibly non-transitive elementary submodel of some $H_\theta$. You said to be generic means to meet every dense subset $D\subset \mathbb{P}$ with $D\in M$, but this is not the right definition. You want to say instead that $G$ meets every such dense set $D$ inside $M$. That is, that $G\cap D\cap M\neq\emptyset$. If we only have $G\cap D\neq\emptyset$, then $M$ will not have access to the conditions $p\in G\cap D$ that are useful when a filter meets a dense set. So it is the corrected definition that treats $\langle M,{\in^M}\rangle$ as a model of set theory in its own right, insisting that for every dense set in this structure, the filter meets it.

Lastly, let me point out that one doesn't need countable models in order to undertake the forcing construction, and one can speak of the forcing extensions of any model of set theory, whether it is countable, transitive, uncountable, nonstandard, whatever. The most illuminating way to do this is via Boolean-valued models, and by taking the quotient, one arrives at the Boolean ultrapower construction. The basic situation is the if $V$ is a model of set theory containing a complete Boolean algebra $\mathbb{B}$, and $U\subset\mathbb{B}$ is an ultrafilter ($U\in V$ is completely fine), then one may form the quotient $V^{\mathbb{B}}/U$ of the $\mathbb{B}$-valued structure, and this is realized as a forcing extension of its ground model $\check V_U$, and furthermore there is an elementary embedding of $V$ into $\check V_U$, called the Boolean ultrapower map. So the entire composition $$V\prec \check V_U\subset \check V_U[G]=V^{\mathbb{B}}/U$$ lives inside $V$. There is no need for $V$ to be countable and no need for $U$ to be generic in any sense, yet $G$, which is the equivalence class of the name $\dot G$ by $U$, is still nevertheless $\check V_U$-generic. You can find fuller details in my paper with D. Seabold, Boolean ultrapowers as large cardinal embeddings.

Lastly, let me point out that one doesn't need countable models in order to undertake the forcing construction, and one can speak of the forcing extensions of any model of set theory, whether it is countable, transitive, uncountable, nonstandard, whatever. The most illuminating way to do this is via Boolean-valued models, and by taking the quotient, one arrives at the Boolean ultrapower construction. The basic situation is the if $V$ is a model of set theory containing a complete Boolean algebra $\mathbb{B}$, and $U\subset\mathbb{B}$ is an ultrafilter ($U\in V$ is completely fine), then one may form the quotient $V^{\mathbb{B}}/U$ of the $\mathbb{B}$-valued structure, and this is realized as a forcing extension of its ground model $\check V_U$, and furthermore there is an elementary embedding of $V$ into $\check V_U$, called the Boolean ultrapower map. So the entire composition $$V\overset{\prec}{\scriptsize\sim} \check V_U\subset \check V_U[G]=V^{\mathbb{B}}/U$$ lives inside $V$. There is no need for $V$ to be countable and no need for $U$ to be generic in any sense, yet $G$, which is the equivalence class of the name $\dot G$ by $U$, is still nevertheless $\check V_U$-generic. You can find fuller details in my paper with D. Seabold, Boolean ultrapowers as large cardinal embeddings.

In regard to question 2, of course we want $G$ to be $H_\theta$-generic, since without this it is easy to make counterexamples to $M[G]\prec H_\theta[G]$. For example, if $M$ is countable we can easily find $M$-generic filters $G$ with $G\in H_\theta$, and in this case, if the forcing is nontrivial then $M[G]$ is definitely not an elementary substructure of $H_\theta[G]=H_\theta$.

In regard to question 2, of course we want $G$ to be $H_\theta$-generic, since without this it is easy to make counterexamples to $M[G]\prec H_\theta[G]$. For example, if $M$ is countable we can easily find $M$-generic filters $G$ with $G\in H_\theta$, and in this case, if the forcing is nontrivial then $M[G]$ is definitely not an elementary substructure of $H_\theta[G]=H_\theta$. This is the argument of your last paragraph, and that is totally right; so the conclusion is that for this question we want to assume $G$ is $V$-generic.