I've read a few different presentations of Cohen's proof. All of them (that I've seen) eventually make a move where a Cartesian product (call it CP) between the (M-form of) $\aleph_2$ and $\aleph_0$ into {1, 0} is imagined. From what I gather, whether $CP \in M$ is what determines whether $\neg$CH holds in M or not such that if $CP \in M$ then $\neg$CH.
Anyway, my question is: Why does this product, CP, play this role? How does it show us that $\aleph_2 \in M$ (the 'relativized' form of $\aleph_2$, not the $\aleph_2 \in V$)? Could not some other set-theoretical object play the same role?
I'm not exactly certain what you are asking, but I'll take a stab anyway.
The basic idea of Cohen's proof (or at least modern interpretations of it) are to adjoin to the ground model a family of $\aleph_2^M$ new reals. Since a real is nothing more than a function $\omega \to 2$, a family of $\aleph_2^M$-many reals may be thought of as a function $\omega_2^V \times \omega \to 2$. (Recall that there is a natural bijection between $C^{A \times B}$ and $( C^B )^A$, and so a function $f : \omega_2^M \times \omega \to 2$ may be thought of as an indexing of $\omega_2^M$-many reals $\{ f_\alpha : \alpha < \omega_2^M \}$ where $f_\alpha ( n ) = f ( \alpha , n )$.
If we can generate such a function, we are left with two very important steps:
(Of course, that $\aleph_2$ is used is not strictly important, and the methodology of the proof goes through for "almost all" uncountable ordinals.)