Denote $[n]=\{1,\dotsc,n\}\subseteq \mathbb{N}$. Consider the finite sum $\sum_{(i,j)\in A} f(x_i,x_j)$ where $A\subseteq [n]^2$. Assume that $A=\cup_{l=1}^n A_l$ is a disjoint union and $A_l=\{(l,j): (l,j)\in A\},l=1,\dotsc,n$. Can I conclude that $$\sum_{(i,j)\in A} f(x_i,x_j)=\sum_{l=1}^n\sum_{(l,j)\in A_l} f(x_i,x_j) ?$$
I think this is the case. But it is not clear to me how to argue that. Maybe I have to see this as $$\sum_{(i,j)\in \cup_{l=1}^n A_l}f(x_i,x_j)=\sum_{(i,j)\in A_1}f(x_i,x_j)+\dotsc +\sum_{(i,j)\in A_n}f(x_i,x_j)\\ =\sum_{(1,j)\in A_1}f(x_i,x_j)+\dotsc +\sum_{(n,j)\in A_n}f(x_i,x_j)\\ =\sum_{l=1}^n\sum_{(l,j)\in A_l} f(x_i,x_j)$$.
What do you think?
If the co-domain of $f$ has associative and commutative addition, then you can reorder the finite terms of the sum as you like.
Note that there seems to be a clerical error in your formulation, in
$$\sum_{l=1}^n\sum_{(\color{red}l,j)\in A_l} f(x_i,x_j)$$
the $\color{red}l$ should be replaced with an $i$.
Or maybe you started with the special case that $A_l=\{(l,j) \in A\}$, which would explain both that error as well as ajotaxte's comment.