If $i \xrightarrow{g} j \xrightarrow{f} k$ in $I$, then
$F(i) \xrightarrow{F(g)} F(j) \xrightarrow{F(f)} F(k) \\ G(i) \xrightarrow{G(g)} G(j) \xrightarrow{G(f)} G(k)$
in $A$. And since $A$ is additive, we have $F(i) \times G(i)$ existing in $A$ and thus:
$F(i) \times G(i) \xrightarrow{\pi^F_i} F(i) \to F(j) \to F(k) \\ F(i) \times G(i) \xrightarrow{\pi^G_i} G(i) \to G(j) \to G(k) $
So that considering the product $F(k) \times G(k)$, there is a unique map $\alpha : F(i) \times G(i) \to F(k) \times G(k)$ such that:
$$ \pi^F_k \alpha = F(f) \circ F(g) \circ \pi_i^F \\ \pi^G_k \alpha = G(f) \circ G(g) \circ \pi_i^G $$
But how do I get that $\alpha(f\circ g) = (F\times G)(f \circ g) = (F \times G)(f) \circ (F\times G)(g)$?
As you said, $\alpha$ is the unique map having the properties you mentioned. So, you just have to show that both $(F\times G)(f \circ g)$ and $(F \times G)(f) \circ (F\times G)(g)$ have those properties to conclude that they are equal.
By definition, $(F\times G)(f \circ g)$ satisfy these properties. For $(F \times G)(f) \circ (F\times G)(g)$, we have $$\pi_k^F \circ (F \times G)(f) \circ (F\times G)(g)=F(f) \circ \pi_j^F \circ (F\times G)(g)=F(f) \circ F(g) \circ \pi_i^F$$ and similarly for $G$.