Let $\mathcal{C}$ and $\mathcal{D}$ two additive categories. Let $F: \mathcal{C} \rightarrow \mathcal{D}$ a functor that $F(A \times B) \simeq F(A) \times F(B)$ for objects $A,B \in \mathcal{C}$. Consider the products $(A\times A, \pi^{A}_1,\pi^{A}_2)$ and $(F(A) \times F(A), \pi^{F(A)}_1,\pi^{F(A)}_2)$. If we denote $f_A: F(A \times A) \rightarrow F(A) \times F(A)$ the isomorphism i'm trying to show that $F(\pi^{A}_i) \circ (f_A)^{-1} = \pi^{F(A)}_i$ for $i = 1,2$.
First, is this true? if it is, i'm confused what diagrams to use.
When saying that a functor $F:{\mathcal C}\to{\mathcal D}$ preserves the product $$(A\times B, \pi_A: A\times B\to A, \pi_B: A\times B\to B)$$ it is usually meant that the triple $$(F(A\times B), F(\pi_A): F(A\times B)\to F(A), F(\pi_B): F(A\times B)\to F(B))$$ is a product of $F(A)$ and $F(B)$.
Your statement then follows from the uniqueness up to unique isomorphism of products:
If $(C,p_1: C\to A, p_2: C\to B)$ and $(D, q_1: D\to A, q_2: D\to B)$ are two products of $A$ and $B$, then there are unique maps $f: C\to D$ and $g: D\to C$ with the properties that $q_i\circ f = p_i$ and $p_i\circ g = q_i$, and these maps are mutually inverse isomorphisms. In particular, if $f: C\to D$ is some map satisfying $q_i\circ f = p_i$, $f$ must be an isomorphism and $f^{-1} = g$, in particular $p_i \circ f^{-1} = q_i$. Applying this to your comparison morphism $f := f_A: F(A\times B)\to F(A)\times F(B)$ between the product $(F(A\times B),F(\pi_A),F(\pi_B))$ and any other product $(F(A)\times F(B),\pi_{F(A)},\pi_{F(B)})$ gives $F(\pi_{A/B}) \circ f_A^{-1} = \pi_{F(A)/F(B)}$ as required.