Let $R$ be a ring with 1, $A$ a right $R$-module and $B$ a left $R$-module. To construct their tensor product, let $F$ be the free abelian group with basis $A\times B$ and $S$ the subgroup generated by all elements of the following three types : $(a,b+b')-(a,b)-(a,b'); (a+a',b)-(a,b)-(a',b); (ar,b)-(a,rb)$. Then $A\otimes_R B=F/S$.
Let $a\otimes b=0$. Then $(a,b)\in S$. So $(a,b)$ can be written as $\sum\limits_{k}\pm(a_k,b_k),$ where $a_k\in A, b_k\in B$ and $a_k$ are $a,a+a',ar$ and $b_k$ are $b,b+b',rb$.
I was wondering if the last sentence is correct.
Since $(a,b) \in S$ it can be written in such a way but you also have the fact that $A \times B$ is a basis of the free abelian group $F$. Thus since $(a,b) \in A \times B$ it is already uniquely written in this basis. Hence every element of your sum must have its opposite somewhere in the sum too, except for one which would be $(a,b)$.
A word of warning, this does not mean that $a=0$ or $b=0$. For example if there is some torsion in the module or some zero divisor in the ring: $$(ra,b)=(ra,b)-(a,rb) + (a,0) -(0\times a,0) \in S \ \text{ if } rb=0$$
You can find a more complete answer (with a caracterisation and some examples) about this warning here.