For a definition of additive category we have 12.3.8 of the following link:
https://stacks.math.columbia.edu/tag/09SE
Let $\mathcal{A}$ an additve category, $f,g \in Hom_{\mathcal{A}}(X,Y)$ and $S \subset Mor(\mathcal {A})$ which is closed under compostions and contains identity morphism for every object in $\mathcal{A}$, the following two properties are equivalent:
Property (1)
These two senteces are equivalent:
$(a)$ There is a morphism $w:Y \to Y'$ in $S$ such $wf=wg$.
$(b)$ There is a morphism $v:X' \to X$ in $S$ such $fv=gv$.
Property (2)
These two senteces are equivalent:
$(i)$ If $wf=0$ for some $w:Y \to Y'$ in $S$, then there is a morphism $v:X' \to X$ in $S$ such $fv=0$.
$(ii)$ If $fv=0$ for some $v:X' \to X$ in $S$, then there is a morphism $w:Y \to Y'$ in $S$ such $wf=0$.
Let me show my attempt of P(2) $\Rightarrow$ P(1). Lets suppose P(2), that means $(i) \iff (ii)$ and I want to prove $(a) \iff (b)$. Lets start proving $(a) \implies (b)$. As we suppose $(a)$ we have that there is a morphism $w:Y \to Y'$ in $S$ such $wf=wg$ and as $\mathcal{A}$ is additive we have that $w(f-g)=0$ IM NOT SURE AT THIS STEP but as we suppose P(2) there is $v:X' \to X$ in $S$ such $(f-g)v=0$ so we got $(b)$. I think this is wrong as we dont use he equivalence per se. For P(1) $\Rightarrow$ P(2) I didnt come with any idea. Aprecciate your help to solve this one.
I'm going to assume that the sentences are each prefixed by for all $f,g\in \mathcal{A}(X,Y)$, otherwise, I think this is false, since (i) and (ii) don't mention the morphism $g$ from (a) and (b).
Observation:
(i) is equivalent to (a) $\implies$ (b), and (ii) is equivalent to (b) $\implies$ (a).
Therefore (i) iff (ii) is equivalent to (a $\implies$ b) iff (b $\implies$ a), which is equivalent to (a $\iff$ b) or (not $b$ and $a$ and not $a$ and $b$), and the latter is false, so this is equivalent to a $\iff$ b, as required.