Understanding the definition of tensor product as a quotient of a free abelian group I've been give the Definition:


  Let F be a free abelian group with a basis $X$ such that.
  $$F = \langle A\times B\mid \emptyset \rangle $$ 
  
  Let $f$ be a subgroup of $F$ generated by the elements $$(a + b, g) - (a,g) - (b,g), \\  n(a,g) - (na,g),(a,g + h) - (a,g) - (a,h), \\ n(a,g) - (a,ng)\\  \forall n \in \mathbb{Z}\space,\forall a,b \in A\space,\forall g,h \in B$$


Then $A\otimes B = F/f$

By problem is I don't understand how the elements stated in the definition form basis for the free abelian group. Could someone provide me with some examples to help understand the group that is being quotient out please.
This bothered me too at first. The key thing to note is that the group operation in $F$ has nothing at all to do with the group operation in $A\times B$. For example, let $A=\mathbb{Z}_2$ and $B=\mathbb{Z}_4$. Then an element of $F$ looks like
$$a_1(0,0)+a_2(1,0)+a_3(0,1)+a_4(1,1)+a_5(0,2)+a_6(1,2)+a_7(0,3)+a_8(1,3)$$
where the $a_i$ are integers. $F$ is the set of all formal integer linear combinations of the elements of $A\times B$. When we take the quotient, we impose relations on these elements. An example of one of the elements that generate the kernel $f$ is
$$(1,1+2)-(1,1)-(1,2)$$
This is sent to $0$ when we take the quotient, which means that
$$[(1,1+2)]-[(1,1)]-[(1,2)]=0$$
or
$$[(1,1+2)]=[(1,1)]+[(1,2)]$$
In tensor product notation we are saying that $1\otimes (1+2)=1\otimes 1+1\otimes 2$, as $(a,b)\mapsto a\otimes b$.
If you write $F=&lt;A\times B|\emptyset&gt;$ then this is just the same as saying that $F$ is the free abelian group generated by the set $A\times B$ (by definition of a presentation). I think what confuses you is a notation failure: it is intended that $X=A\times B$. Just forget about the $X$. 

In any case, if you want to understand the tensor product, you should distinguish conceptually between the definition (universal property) and the proof of existence (which is the $F/f$). The former is by what the tensor product does, and the second one you can essentially forget about, you never use it in dealing with tensor products (I say this as someone who uses every day, it is just not useful in practice); in fact you can find this statement in the standard textbook on commutative algebra by M. Atiyah.

To understand this group $F$: let $X$ be any set (in your case you have $X:=A\times B$). We can form the set of maps $X\rightarrow\mathbf{Z}$ with finite support, and define the free abelian group $F(X)$ generated by $X$ to be this set with the pointwise defined group operation. Note that the elements $\delta_x$ ($x\in X$) given by $\delta_x:X\rightarrow\mathbf{Z}$, $\delta_x(y)=0$ for $x\neq y$ and $\delta_x(x)=1$ form a basis of this group: for $\alpha:X\rightarrow\mathbf{Z}$ you have $\alpha=\sum_{x\in X}\alpha(x)\delta_x$ and this expression is unique. The map from $X$ to $F(X)$ by $x\mapsto\delta_x$ is injective, and if we identify $X$ with its image, then $X$ can be viewed as a basis for this group.

Do you still need an example? I think you are mostly confused about notation. In any case: take for instance $X=\mathbf{N}$. Then $F(X)$ is just the set of sequences in $\mathbf{Z}$ with finite support (meaning that for every sequence only finitely many members are nonzero).