Let $M,N$ be $R$-modules. Let $C$ be the $R$-module of formal linear combinations of the form $$\displaystyle\sum_{i=1}^nr_i(x_i,y_i)$$ whit coefficients $r_i$ in $R$ and $(x_i,y_i)$ in $M\times N$. Let $D$ be the submodule of $C$ generated by the elements of $C$ of the form $$(x+x',y)-(x,y)-(x',y)$$ $$(x,y+y')-(x,y)-(x,y')$$ $$(ax,y)-a(x,y)$$ $$(x,ay)-a(x,y)$$ Let $T$ be the quotient module $C/D$. Let $g\colon M\times N\to T$ be the map sending a pair $(x,y)$ to the class of $(x,y)$ modulo $D$. Then $g$ is bilinear and the pair $(T,G)$ satisfies the universal property of tensor product, i.e every bilinear map from $M\times N$ to any $R$-module $P$ factors through $T$.
What I can't understand is why $(x,y)$ is an element of $C$. I mean, $C$ is the set of formal linear combinations as above, so $1$ times $(x,y)$ is an element of $C$. If $I$ say that $1\cdot (x,y)=(x,y)$, then I don't understand what "formal" means. Is $(x,y)$ just an abbreviation for $1\cdot (x,y)$?
What I think is that "formal" stays for: $C$ is the direct sum $$\displaystyle\oplus_{(x,y)\in M\times N} R(x,y)$$ Here $R(x,y)$ is the set of formal multiples of $(x,y)$ with coefficients in $R$, which is an $R$-module with the obviuos multiplication. But an element of $R$ should always appear. So I can't understand how $(x,y)$ (or $(x+x', y)$ or $(x, y+y')$ etcetera) can be elements of that direct sum.
If $X$ is a set and $F(X)$ is a free algebraic structure generated by $X$ (e.g. the case at hand: $C$ is the free $R$-module generated by $M \times N$), there is a canonical function from $X$ to the set of elements of $F(X)$.
That is, there is a canonical way to interpret a generator $x \in X$ as referring to an element of $F(X)$. It is standard to use this interpretation whenever an element of $X$ is written in a place where an element of $F(X)$ is expected. Sometimes a mild decoration is added, such as surrounding the element with brackets, e.g. $[x]$.
So yes, $(x,y)$ is being used to refer to the element of that direct sum that has $1$ in the $(x,y)$-indexed component, and a $0$ in every other component.
Note, incidentally, that you were using a similar sort of convention when you were interpreting the element $1 \cdot (x,y) \in R(x,y)$ as being an element of the direct sum.