Is this computation of the tensor product correct? I'm reading the proof of the existence of the tensor product. If $M,N$ are two $R$-modules then we can construct the tensor product $T$ as the quotient $C/D$ where $C$ is the free module over $M \times N$ and $D$ is the submodule generated by the set of all elements in $C$ of the form 
$$(m+m^\prime, n) - (m,n) - (m^\prime, n)$$
$$ (m, n+n^\prime) -(m,n) - (m,n^\prime) $$
$$ (am, n) - a(m,n)$$
$$ (m,an) - a(m,n)$$
I use $(m,n)$ to denote the element $e_{(m,n)} \in F(M\times N)$. Since $F(S) \cong \bigoplus_{s \in S} R$ I picture these elements as $e_{(m,0)} = (0, \dots, 0,1, 0, \dots)$ where the $1$ here is at position $m$ and  $e_{(m,n)}$ the sequence with $1$ at position $m \cdot |M| + n$ and so forth.
Is this correct so far?
Now I wanted to see what this looks like. So I computed the tensor product of $M = N = \mathbb Z / 2 \mathbb Z$ over $R=\mathbb Z$. For $C$ I get that $C \cong \mathbb Z^4$. Then I computed all the elements above and noticed that $D \cong \langle \{(1,0,0,0), (0,1,0,0), (0,0,1,0)\} \rangle$. Hence 
$$M \otimes N = \mathbb Z / 2 \mathbb Z \otimes_{\mathbb Z} \mathbb Z / 2 \mathbb Z = \langle (0,0,0,1) \rangle \cong \mathbb Z$$
Is this correct? 
 A: Your  last isomorphism above cannot hold; the guy on the left is a finite group but the one on the right is 100% not a finite group!
What it should be isomorphic to is $\Bbb{Z}/d\Bbb{Z}$ where $d$ is the greatest common divisor of $2$ and $2$, in this case $2$ itself so that $\Bbb{Z}/2\Bbb{Z} \otimes_\Bbb{Z} \Bbb{Z}/2\Bbb{Z} \cong  \Bbb{Z}/2\Bbb{Z}$ . To see this, given any elementary tensor $a \otimes b$ in the tensor product, there are only 4 possibilities: $a$ odd $b$ even, $a$ odd $b$ odd, $a$ even $b$ odd, $ a$ odd $b$ even. But the cases where you have an even appearing are just zero because 
$$\begin{eqnarray*} 0 \otimes 1  &=& (0 + 0) \otimes 1 \\
&=& 0 \otimes 1 + 0 \otimes 1 \\
\implies 0 \otimes 1 &=& 0 \end{eqnarray*} $$
and similarly $1 \otimes 0 = 0$.   Hence there are only two distinct elements that appear, namely $1 \otimes 1$ and  $0$ so that your tensor product is isomorphic to the cyclic group of order 2. 
A: I will add to the comment I gave beneath your question. I know you said that you were interested in the specific construction, but perhaps later you will find this answer useful too.
We know that for any ideal $I \lhd R$, and any $R$-module $M$, we have the isomorphism:
$$ R/I \otimes_R M \cong M/IM$$
To see this, take the exact sequence $$0 \to I \to R \to R/I \to 0$$
and tensor by $M$ (recalling that tensoring by $M$ preserves right-exact sequences), then use the standard isomorphism theorem for modules.
So in your case it follows that
$$ \mathbb{Z}/2 \mathbb Z  \otimes_\mathbb{Z} \mathbb{Z}/2 \mathbb Z \cong \frac{\mathbb{Z}/2\mathbb Z }{ \langle 2 \rangle \mathbb{Z} / 2\mathbb{Z}} \cong \mathbb{Z} / 2 \mathbb{Z} $$
Since the bottom part of the quotient is just $0$.
