Proposition 3.1.15 (Multiplication). Let $A$, $B$ be C*-algebra, the tensor product $A\odot B$ (denotes the algebraic tensor product) has a multiplication defined by $$(\sum\limits_{i}a_{i}\otimes b_{i})(\sum\limits_{j}c_{j}\otimes d_{j})=\sum\limits_{i,j}a_{i}c_{j}\otimes b_{i}d_{j}.$$
To prove that this multiplication works, we first consider $L(A\odot B)$, the vector space of all linear maps $A\odot B\rightarrow A\odot B$. If $M_{a}: A\rightarrow A$ is left multiplication by $a\in A$ and $M_{b}: B\rightarrow B$ is left multiplication by $b\in B$, then, for every pair $(a, b)\in A \times B$ we have the tensor product map $M_{a}\otimes M_{b}\in L(A\odot B)$. It is routine to check that $$A\times B\rightarrow L(A\odot B),~(a, b)\mapsto M_{a}\otimes M_{b}$$ is a bilinear map. By universality there is a linear map $M: A\odot B \rightarrow L(A\odot B )$ such that $M(a\otimes b)=M_{a}\otimes M_{b}$. Finally one checks that the bilinear map $$A\odot B \times A\odot B \rightarrow A\odot B , (x, y)\mapsto M(x)y$$ defines our multiplication.
It is a quotation above. I do not know what is the meaning of "to prove this multiplication works". Does the author want to show the multiplication $\sum\limits_{i,j}a_{i}c_{j}\otimes b_{i}d_{j}\in A\odot B$?
The multiplication is defined in terms of the individual elements of $\sum_ia_i\otimes b_i$ and $\sum_ic_i\otimes d_i$. This means that you need it to be independent of the choice; that is, if $\sum_ia_i'\otimes b_i'=\sum_ia_i\otimes b_i$ and $\sum_ic_i'\otimes d_i'=\sum_i c_i\otimes d_i$, you need $\sum_ia_i'c_i'\otimes b_i'd_i'=\sum_ia_ic_i\otimes b_i d_i$.