If a pointed variety is a subtractive category, then it is a subtractive variety. Let $\mathcal{V}$ be a variety of algebras. Every class of algebras is a category. In particular, assume that $\mathcal{V}$ is a subtractive category. This means that in $\mathcal{V}$ every reflexive left punctual relation is also right punctual. Assume also that $\mathcal{V}$ is pointed, that is that V has a constant 0. (A relation $R$ on $X$ is left punctual if the left projection of $R$ on $X$ has a section, similarly for right punctuality)
I want to prove that, under assumptions above, $\mathcal{V}$ is a subtractive variety: this means that $\mathcal{V}$ has a binary term $s$ satisfying for every variable $x$, the identities $s(x,x)=0$ and $s(x,0)=x$.
Let’s start with the proof. Let $F(x)$ be the free algebra over the set $\{x\}$ of variables. Let $R$ be the relation on $F(x)$ homomorphically generated by the subset $\{(x,0),(x,x)\}$ of $F(x)\times F(x)$. Since $(x,x)$ is in $R$, $R$ is reflexive. Since $(x,0)$ is in $R$, $R$ is left punctual. Since $\mathcal{V}$ is a subtractive category, $R$ is also right punctual, hence $(0,x)$ is in $R$. Then there exists a binary term $s$ such that $s( (x,x), (x,0) )= (0,x)$. Writing the identity component-wise, we get $s(x,x)=0$ and $s(x,0)=x$. This concludes the proof.
Now, what I couldn’t understand:
1) Why does $(x,x)\in R$ imply $R$ reflexive? Shouldn’t I prove that $(a,a)$ is in $R$ for every $a$ in $F(x)$, not only for $x$?
2) How do I know that $F(x)$ is in $\mathcal{V}$?
3) Why does $(0,x)$ is realized as a binary term $s$ applied to $(x,x) $and $(x,0)$. How do I know it is binary?
4) What does it mean ‘component-wise’ at the end of the proof? I mean, I know the meaning of ‘component-wise’, but in the actual case, what I am actually doing when writing that identity component-wise and why is it possible to do that?
 A: (0) The definition of left punctual is not right. The correct definition makes sense for any variety with $0$, and it is: a relation $R$ on $X$ is left punctual if the first projection $\pi_1:R\to X$ has a section $\sigma:X\to R$ whose composition with second projection $\pi_2\sigma:X\to X$ is the constant zero map. 
(1) Why does $(x,x)\in R$ imply $R$ reflexive? Shouldn’t I prove that $(a,a)$ is in $R$ for every $a$ in $F(x)$, not only for $x$? $R$ is a subalgebra of $X\times X$, for $X=F(x)$. Since $x$ generates $X$, the pair $(x,x)$ generates the diagonal of $X\times X$. Thus if $(x,x)\in R$, it follows that the diagonal is contained in $R$ as well, and so $R$ is reflexive.
(2) How do I know that $F(x)$ is in $\mathcal{V}$? This is a property of varieties. One common definition of a variety is: a class closed under the formation of homomorphic images, subalgebras, and products. But one can construct free algebras as subalgebras of products, so one only needs closure under subalgebras and products to guarantee that a class contains free algebras. 
(3) Why does $(0,x)$ is realized as a binary term $s$ applied to $(x,x) $and $(x,0)$. How do I know it is binary? The corrected definition of subtractive (see item (0)) guarantees that the relation generated by $(x,x)$ and $(x,0)$ contains the pair $(0,x)$. Now the following are equivalent: (i) $(0,x)\in \langle (x,x), (x,0)\rangle$, and (ii) there is a binary term operation $s$ such that $s((x,x),(x,0))=(0,x)$. One can prove this by showing, e.g., that in general 
$\langle a, b\rangle = \{s(a,b)\;|\;s\;\textrm{a binary term operation}\}$.
(4) What does it mean ‘component-wise’ at the end of the proof? I mean, I know the meaning of ‘component-wise’, but in the actual case, what I am actually doing when writing that identity component-wise and why is it possible to do that? Not sure I understand the question, but let me give it a shot. The relation $R$ is the subalgebra of $X\times X$ generated by $(x,x)$ and $(x,0)$. $R$ contains $(0,x)$, so there is a binary term operation $s$ such that $s((x,x),(x,0))=(s(x,x),s(x,0))=(0,x)$. We are computing in $X\times X$, which consists of ordered pairs, and term operations are evaluated on pairs coordinatewise: $s((a,b),(c,d))=(s(a,c),s(b,d))$. You ask why we evaluate term operations coordinatewise in a product: one could say, That is the definition of the product, or one could say, That is forced by the universal property of products.

Edit 11/20/17.
Do you have some references or hints about what you wrote in the line: "One can prove this by showing, e.g., ..." at point 3) of your answer?
Proposition 13 of these notes is relevant, but I will add detail here.
Let $T(x,y)$ be the term algebra in the variables $x$ and $y$. It is generated by the set $\{x,y\}$. Let $A=\langle a, b\rangle$. $T(x,y)$ is absolutely free over $\{x,y\}$, so there is a unique homomorphism $\varphi\colon T(x,y)\to A\colon x\mapsto a, y\mapsto b$. By necessity, $\varphi(s(x,y))=s(a,b)$, so $\varphi(T(x,y))=\{s(a,b)\;|\;s\;\textrm{a binary term operation}\}$. This explains why $\{s(a,b)\;|\;s\;\textrm{a binary term operation}\}$ is a subalgebra of $A$ (it is an image). Since it contains $a$ and $b$, we derive that $\{s(a,b)\;|\;s\;\textrm{a binary term operation}\}\supseteq \langle a, b\rangle$.
Homomorphisms map generating sets to generating sets, so we also get that $\varphi(T(x,y))=\{s(a,b)\;|\;s\;\textrm{a binary term operation}\}$ is generated by $\varphi(\{x,y\})=\{a,b\}$, thereby showing that 
$\{s(a,b)\;|\;s\;\textrm{a binary term operation}\} = \langle a, b\rangle$.
