Given an abelian category $\mathbb{A}$.
There are numerous definitions of zero morphism:
For every pair of objects $X,Y\in\mathbb{A}$
there is a morphism $0_{YX}:X\to Y$ such that..
Factorization
.. it factors through a zero object $0_{YX}=0_{Y0}0_{0X}$.
Composition
..it annihilates morhphisms $0_{YX}f_{XA}=0_{YA}$ and $g_{BY}0_{YX}=0_{BX}$.
Addition
..it acts as the identity $f_{YX}+0_{YX}=f_{YX}$.
Now are these zero morphisms the same?
On
Yes they are, and in fact I would propose a third description which needs no further existence proof and allows for quick verification of the properties that you mentioned:
Definition: For any two $X,Y$, the zero morphism is the morphism $X\to 0\to Y$, where the morphisms $X\to 0$ and $Y\to 0$ used are the unique morphisms (since $0$ is both terminal and initial).
Starting from this, do you want to try to show that it implies your properties?
Fix a zero object $0$. It's easy to prove that the composition $X\to0\to Y$ is independent of the zero object (because if $0'$ is another one there is a unique isomorphism $0\to0'$).
Let's take as definition of zero morphism $X\to Y$ one that factors through the chosen zero object. Such a morphism is unique (easy proof), so we can denote it by $0_{XY}$.
Consider now $f\colon X\to Y$ and do the composition $0_{YZ}f$; then we have $$ X\xrightarrow{f}Y\mathrel{\underbrace{\to 0\to}_{0_{YZ}}}Z $$ so $0_{YZ}f$ factors through the zero object; hence $0_{YZ}f=0_{XZ}$.
Conversely, if $Y\xrightarrow{z}Z$ has the property that $zf=0$, for all $f$ with codomain $Y$, then, in particular it does for $Y\xrightarrow{1}Y$, so $z$ must factor through the zero object.
Similarly (or by duality) on the other side.
In order to show that $0_{XY}$ acts as the neutral element for sum of morphisms, recall how sum of morphisms is defined (or characterized) via a suitable diagram.