I found two different definitions of 'subtractive category' and I am wondering if they are the same or what.
1) A category is subtractive if it is pointed, regular and for every object $X$ of the category, the morphism $p_X$ is a regular epimorphism
2) A category is subtractive if it is pointed, regular and for every object $X$ of the category, every reflexive left puctual relation on $X$ is also right-punctual.
Explanation: the morphism $p_X$ is the following composite: $$p_X= \binom{1_x}{0}\circ\delta_X:D(X)\to X$$
where $(D(X),\delta_X)$ is the kernel of the codiagonal morphism $\nabla_X=\binom{1_X}{1_X}:X+X\to X$ and $\binom{1_X}{0}:X+X\to X$ the factorization through $X+X$ of the pair of morphisms $\{1_X,0\}$.
Given a reflexive relation $r:R\to X$ on $X$, with $r=(r_1,r_2)$, being left puctual means that there exists a morphism $s:X\to R$ such that $r_1s=1_X, r_2s=0$. Being right-puntual means that there is a morphism $t:X\to R$ such that $r_2t=1_x$ and $r_1t=0$.
Indeed, the two definitions are equivalent (assuming the category has coproducts).
Assume first that 1) is true. Given a left punctual reflexive relation $r=(r_1,r_2)$, then the universal property of the coproduct yields a map $\binom{\delta}{s}:X+X\to R$ (where $\delta$ is the arrow $X\to R$ given by the reflexivity, and one can check that $$r_1\binom{\delta}{s}=\binom{r_1\delta}{r_1s}=\binom{1_X}{1_X}=\nabla_X$$ and $$r_2\binom{\delta}{s}=\binom{r_2\delta}{r_2s}=\binom{1_X}{0}.$$ The first equation implies that $\binom{\delta}{s}$ must restrict to a map $\theta:D(X)\to Ker(r_1)$ between the kernels. Then $$r_2\circ \ker(r_1) \circ \theta=r_2\binom{\delta}{s}\delta_X=\binom{1_X}{0}\delta_X=p_X$$ is a regular epimorphism, and thus $r_2\circ \ker(r_1)$ is also a regular epimorphism. But it is also a monomorphism: indeed, for any arrows $u,v$ one has $$r_1\circ \ker(r_1)\circ u=0=r_1\circ \ker(r_1)\circ v,$$ so if $$r_2\circ \ker(r_1)\circ u=r_2\circ \ker(r_1)\circ v,$$ one must have $\ker(r_1)\circ u=\ker(r_1)\circ v$ and thus $u=v$. Thus $r_2\circ \ker(r_1)$ is an isomorphism, and we can assume WLOG that is actually equal to $1_X$, which shows that $r$ is right punctual.
Now assume that 2) is true. Consider the image factorization $rp$ of the canonical arrow $$\left(\binom{1_X}{1_X},\binom{1_X}{10}\right):X+X\to X\times X.$$ Then $r$ is a monomorphism $R\to X\times X$, thus a relation; and it is reflexive and left punctual (thanks to the coproduct injections $i_1$ and $i_2$ composed with $p$), hence also right punctual. Then the arrow $t$ such that $r_1t=0$ and $r_2t=1_X$ must factor through $\ker(r_1)$, which means that $r_2\circ \ker(r_1)$ is a split epimorphism; since it is also a mono, it is an isomorphism, and then the factorization of $t$ through $Ker(r_1)$ must also be an isomorphism, so we can assume WLOG that $t=\ker(r_1)$. Now the identity $r_1p=\nabla_X$ gives once again a restriction $\theta$ of $p$ to the kernels. Moreover, one can prove that the commutative square $p\delta_X=\ker(r_1)\theta$ is actually a pullback, thus $\theta$ is a regular epimorphism. Thus $$p_X=\binom{1_X}{0}\delta_X=r_2 p\delta_X=r_2\ker(r_1)\theta=\theta$$ is a regular epimorphism.