Q:

About Semigroup & Semiring homomorphism

This is Theorem 2.1.1 from Scharlau's book Quadratic and Hermitian Forms

Can somebody explain me why $R$ has zero element $[a,a]$ and negative element $[b,a]$? My guess: I think it is due to the equivalence relation we define we want $c=0$ element hence negative of $[a,b]$ is $[b,a]$.
Why it's a homomorphism is clear but why $[a,b]=[a,0]-[b,0]$? Also, why $f'$ is a well defined group homomorphism and why it is uniquely determined is not clear to me. Somebody help regarding this. 

A:

why R has zero element $[a,a]$

In order to verify that some element "$E$" is the zero you have to check that $a+E=E+a=a$ for all $a$. So let's do it. Let $a, r, q$ be any elements of $R^+$. Then
$$[a,a]+[r,q]=[a+r,a+q]$$
by the definition. Also by the definition of $\sim$ we have $(r,q)\sim(a+r,a+q)$ because you can simply choose $c:=a$. It follows that $[a+r, a+q]=[r,q]$ and thus $[a,a]$ is the neutral element. Note that the element $[a,a]$ does not depend on the choice of $a$, i.e. $[a,a]=[b,b]$ for any $a,b$.

why R has (...) negative element $[b,a]$

Again, you need to follow the definition of the (additive) inverse: $I$ is the inverse of $a$ if $a+I=I+a=E$ where $E$ is the neutral element. We already know that $[a,a]$ is neutral. Now let $r,q$ be two elements and consider
$$[r,q]+[q,r]=[r+q,r+q]$$
by the definition. Now since $[a,a]$ is the neutral element and it doesn't depend on the choice of $a$, then we simply put $a:=r+q$ to obtain that $[r+q,r+q]$ is the neutral element and so $[q,r]$ is the (additive) inverse of $[r, q]$.

why $[a,b]=[a,0]-[b,0]$?

What is "$-$"? It is the operation of taking the additive inverse, i.e. $-[r, q]=[q, r]$ as we've proved earlier. With that you have
$$[a,0]-[b,0]=[a,0]+[0,b]=[a,b]$$
Note the subtlety: this only makes sense if $R$ has zero to begin with. But it's not a problem if it doesn't have because you can always add the neural element to any semisubgroup, simply by forcing $a+0=0+a=a$ for any $a$.

why $f'$is well defined group homomorphism 

You have $f'([a,b])=f(a)-f(b)$. In order to prove that it is well defined we have to show that the definition doesn't depend on the choice of $a,b$. So assume that $[a,b]=[a',b']$, i.e. $a+b'+c=a'+b+c$ for some $c$. Then we have
$$f(a)+f(b')+f(c)=f(a+b'+c)=f(a'+b+c)=f(a')+f(b)+f(c)$$
Note that the codomain of $f$ is an abelian group hence we can subtract $f(c)$ from both sides to obtain
$$f(a)+f(b')=f(a')+f(b)$$
$$f(a)-f(b)=f(a')-f(b')$$
$$f'([a,b])=f'([a',b'])$$

why it is uniquely determine

So assume that $g:R\to S$ is another homomorphism such that $g([r,0])=f(r)$ for all $r\in R^+$. We have to show that $g(x)=f'(x)$ for all $x\in R$. Indeed
$$g([r,q])=g([r,0]-[q,0])=g([r,0])-g([q,0])=f(r)-f(q)=f'([r,q])$$
Side note: The construction is widely known as the Grothendieck group/ring. You can read more about it here: https://en.wikipedia.org/wiki/Grothendieck_group

why it is equivalence relation as I cannot prove symmetry 

Reflexivity: Let $r, q$ be two elements of $R^+$. Pick any $c\in R^+$. Then $r+q+c=r+q+c$ meaning $(r,q)\sim (r,q)$.
Symmetry: Let $r,q,r',q'\in R^+$. Assume that $(r,q)\sim (r',q')$. Then there is $c\in R^+$ such that $r+q'+c=r'+q+c$. Switch sides to obtain $r'+q+c=r+q'+c$ meaning $(r',q')\sim (r,q)$.
Transitivity: Let $r,q,r',q', r'', q''\in R^+$. Assume that $(r,q)\sim (r',q')$ and $(r',q')\sim (r'', q'')$. Then we have $c, c'\in R^+$ such that $r+q'+c=r'+q+c$ and $r'+q''+c'=r''+q'+c'$. Then we have
$$r+q''+(q'+c+c)'=(r+q'+c)+q''+c'=$$
$$=(r'+q+c)+q''+c'=(r'+q''+c')+q+c=$$
$$=(r''+q'+c')+q+c=r''+q+(q'+c+c')$$
meaning $(r,q)\sim (r'', q'')$.

Examples

Let $G$ be an abelian group. Consider $H\subseteq G$ a semisubgroup, i.e. $a+b\in H$ whenever $a,b\in H$. Assume furthermore that for any $g\in G$ there is $h,k\in H$ such that $g=h-k$. Then the Grothendieck construction for $H$ yields $G$ up to isomorphism. Indeed, if $\overline{H}$ is the Grothendieck extension of $H$ then we define $f:\overline{H}\to G$ by $f([h,k])=h-k$. You can easily check that $f$ is a monomorphism (kernel is trivial). The fact that every element of $G$ can be represented as $h-k$ means that $f$ is an epimorphism. All together $f$ is an isomorphism.
So for some concrete examples: let $\mathbb{Z}$ be the group of integers and let 
$$\mathbb{Z}_{\geq k}=\{k,k+1,k+2,\ldots\}=\{n\in\mathbb{Z}\ |\ n\geq k\}$$
You can easily verify that $\mathbb{Z}_{\geq k}$ is a semisubgroup for $k\geq 0$. Furthermore every element $m\in\mathbb{Z}$ can be written as $m=(k+m)-k$ so $\mathbb{Z}_{\geq k}$ satisfies our other condition. Therefore $\overline{\mathbb{Z}_{\geq k}}\simeq\mathbb{Z}$.

about that R is uniquely determined upto isomorphisam.

So assume that we have two groups/rings: $R_1, R_2$, both satisfy the universal property. So we have two homomorphisms: $i_1:R^+\to R_1$ and $i_2:R^+\to R_2$. Now since $R_1$ and $R_2$ have the universal property we can "extend" both of them but with opposite groups to obtain $e_1:R_2\to R_1$ and $e_2:R_1\to R_2$. By the definition $e_2\circ i_1=i_2$ and $e_1\circ i_2=i_1$. Now consider the composition $e_1\circ e_2$. It follows that 
$$e_1\circ e_2\circ i_1=e_1\circ i_2=i_1$$
And since the identity $\text{id}_1:R_1\to R_1$ satisfies the same: $\text{id}_1\circ i_1=i_1$ it follows that $e_1\circ e_2=\text{id}_1$ by the uniquenes of the Grothendieck extension.
Analogously we prove that $e_2\circ e_1=\text{id}_2$ which proves that both $e_1$ and $e_2$ are inverses of each other so $R_1$ is isomorphic to $R_2$.

