Unfortunately I can't check that $\displaystyle 2\otimes 1$ belongs to the denominator, so to speak, of the underlying quotient group, because I don't know how to perform operations on that quotient group. In particular, I don't know how to perform operations on the numerator of that quotient group, which is perhaps the SOURCE of my problem in understanding the definition of a tensor product.

According to my definition, we construct $\displaystyle S\otimes_R N$ by considering the free $\displaystyle \mathbb{Z}$-module over $\displaystyle S\times N$. But $\displaystyle S\times N$ is supposed to be an abelian group in its own right, and I don't understand how.

ANY $\displaystyle \mathbb{Z}-$module, whether free or not, is an abelian group.

The example in page 339 already dealt with this important case
For example, one idea would be to say that since $\displaystyle S$ and $\displaystyle N$ are both abelian groups individually, then we can just carry over the addition operations from each of them so that $\displaystyle (s_1,n_1)+(s_2,n_2)=(s_1+s_2,n_1+n_2)$. Yet clearly this is not the case. But if not that, then how do we perform operations inside the group $\displaystyle S\times N$ ?

Coordinatewise, of course. This is dealt with in page 353.
Here's what my textbook has to say on the subject:

link