I am trying to understand Grothendieck's AB4(*) conditions https://stacks.math.columbia.edu/tag/079A but I don't know how product and coproduct of exact sequences are defined. My only guess is that with regard to morphisms it would be the same as taking a natural transformation between discrete diagrams and then applying the functor $\textrm{(co)}\lim: \mathcal{A}^\mathcal{C} \to \mathcal{A} $ to it, as products and coproduct are instances of limits and colimits respectively.
Anyway, in the finite case, would this correspond to the natural map from coproduct to product that we have in all abelian categories as an application of the universal properties and the isomorphism between products and coproducts?
A functor category ${\cal A}^J$ is abelian if $\cal A$ is abelian, and since limits and colimits are computed pointwise in functor categories, you know what a biproduct, a kernel and a cokernel are in ${\cal A}^J$. So, in ${\cal A}^J$ the notion of exact sequence boils down to "a family, indexed by $J$, of exact sequences".
Let's just consider your case, when $J$ is discrete, i.e. a set: ${\cal A}^J = \prod_{j\in J} \cal A$ is abelian and an exact sequence in ${\cal A}^J$ is precisely an exact sequence $0\to A_j \to B_j\to C_j\to 0$ for each $j\in J$, i.e. a functor $J \to \text{ExSeq}(\cal A)$, of which now you want to compute the co/limit.
This co/limit turns out to be the sequence $0\to \prod A_j \to \prod B_j\to\prod C_j\to 0$ or $0\to \sum A_j \to\sum B_j\to\sum C_j\to 0$, because it can be proved in an elementary fashion that 1. they are exact; 2. they have the correct universal property in the category of exact sequences (regarded as a subcategory of a category of diagrams).
Alternatively, one can prove that $\sum$ and $\prod$ are exact; I think this is equivalent to the fact that the first argument works. (Or at least I'm sure it's true for categories of modules.)