[proofplan] The proof uses exactly the defining bilinearity of composition in a preadditive category. Fixing the morphism $f:A \to B$, postcomposition by $f$ is a group homomorphism on hom-groups, so it sends the additive identity $0_{X,A}$ to the additive identity $0_{X,B}$. Similarly, precomposition by $f$ is a group homomorphism, so it sends $0_{B,Y}$ to $0_{A,Y}$. [/proofplan]

[step:Show that postcomposition by $f$ sends $0_{X,A}$ to $0_{X,B}$] Fix objects $A,B,X \in \mathcal C$ and a morphism $f \in \mathcal C(A,B)$. Define the postcomposition map \begin{align*} P_f:\mathcal C(X,A) &\to \mathcal C(X,B) \\ g &\mapsto f \circ g . \end{align*} Since $\mathcal C$ is preadditive, composition is additive in each variable. Therefore $P_f$ is a homomorphism of abelian groups. The element $0_{X,A}$ is the additive identity in $\mathcal C(X,A)$. Hence \begin{align*} P_f(0_{X,A}) &= P_f(0_{X,A}+0_{X,A}) \\ &= P_f(0_{X,A})+P_f(0_{X,A}). \end{align*} Adding the additive inverse of $P_f(0_{X,A})$ in the abelian group $\mathcal C(X,B)$ to both sides gives \begin{align*} P_f(0_{X,A})=0_{X,B}. \end{align*} By the definition of $P_f$, this is exactly \begin{align*} f\circ 0_{X,A}=0_{X,B}. \end{align*} [/step]

[step:Show that precomposition by $f$ sends $0_{B,Y}$ to $0_{A,Y}$] Fix an object $Y \in \mathcal C$. Define the precomposition map \begin{align*} Q_f:\mathcal C(B,Y) &\to \mathcal C(A,Y) \\ h &\mapsto h \circ f . \end{align*} Since $\mathcal C$ is preadditive, composition is additive in each variable. Therefore $Q_f$ is a homomorphism of abelian groups. The element $0_{B,Y}$ is the additive identity in $\mathcal C(B,Y)$. Thus \begin{align*} Q_f(0_{B,Y}) &= Q_f(0_{B,Y}+0_{B,Y}) \\ &= Q_f(0_{B,Y})+Q_f(0_{B,Y}). \end{align*} Adding the additive inverse of $Q_f(0_{B,Y})$ in the abelian group $\mathcal C(A,Y)$ to both sides gives \begin{align*} Q_f(0_{B,Y})=0_{A,Y}. \end{align*} By the definition of $Q_f$, this is exactly \begin{align*} 0_{B,Y}\circ f=0_{A,Y}. \end{align*} [/step]

[step:Conclude both zero-composition identities] The two preceding steps prove, for arbitrary objects $A,B,X,Y \in \mathcal C$ and arbitrary $f \in \mathcal C(A,B)$, that \begin{align*} f \circ 0_{X,A} &= 0_{X,B}, & 0_{B,Y} \circ f &= 0_{A,Y}. \end{align*} This is precisely the claimed compatibility of zero morphisms with composition in a preadditive category. [/step]