About the group law on the extended square class group 
We define a group law on the set
  $$
  Q(K) := \mathbb{Z}/2\mathbb{Z} \times K^\bullet/K^{\bullet 2}
$$
  as follows
  \begin{align*}
  (0, \alpha) + (0, \beta) &= (0, \alpha \beta) \\
  (1, \alpha) + (0, \beta) &= (1, \alpha \beta) \\
  (1, \alpha) + (1, \beta) &= (0, -\alpha \beta).
\end{align*}
  It is trivial to check that, in fact, this satisfies the axioms of an abelian group.
  $Q(K)$ is called the extended square class group.
  Next, one checks equally trivially that
  $$
  (e,d) \colon W(K) \to Q(K)
$$
  is a group homomorphism.
  Thus, $e$ and $d$ together yield a reasonable invariant.
(Original image here.)

This is excerpt from W. Scharlau: Quadratic and Hermitian Forms,
page 37.
In this $Q(K)$ we define group law as given I don't understand what is third law saying $(1,\alpha) + (1,\beta) = (0,-\alpha\beta)$ is fine in first coordinate since it is from two element group but why $-\alpha \beta$ in second coordinate. Also how it satisfy abelian axioms? What I think is since both groups are abelian there cartesian product is also abelian. And last question is how to show it is homomorphisam? 
 A: Let $G$ be a group containing an element $g$ of the center of $G$.
Let us define a law on the set $K = \mathbb{Z}/2\mathbb{Z} \times G$ by setting
\begin{align}
(0,x)(0,y) &= (0,xy) \\
(0,x)(1,y) &= (1,xy) \\
(1,x)(0,y) &= (1,xy) \\
(1,x)(1,y) &= (0,gxy)
\end{align}
I claim that $K$ is a group for this law. To check associativity, one needs to compare
$A = \bigl((a,x)(b,y)\bigr)(c,z)$ and $B = (a,x)\bigl((b,y)(c,z)\bigr)$. The only nontrivial case occur when at least two elements of the set $\{a, b, c\}$ are equal to $1$. Now, 
\begin{align}
  \bigl((1,x)(1,y)\bigr)(0,z) &= (0,gxy)(0,z) = (0,gxyz) = (1,x)(1,yz)= (1,x)\bigl((1,y)(0,z)\bigr) \\
  \bigl((1,x)(0,y)\bigr)(1,z) &= (1,xy)(1,z) = (0,gxyz) = (1,x)(1,yz)= (1,x)\bigl((0,y)(1,z)\bigr) \\
\bigl((0,x)(1,y)\bigr)(1,z) &= (1,xy)(1,z) = (0,gxyz) = (0,x)(1,yz)= (0,x)\bigl((1,y)(1,z)\bigr) \\
\bigl((1,x)(1,y)\bigr)(1,z) &= (1,gxy)(1,z) = (1,gxyz) = (1,x)(1,yz)= (1,x)\bigl((1,y)(1,z)\bigr)
\end{align}
Clearly $(0,1)$ is the identity element, the inverse of $(0,x)$ is $(0, x^{-1})$ and the inverse of $(1,x)$ is $(1,g^{-1}x^{-1})$. Thus $K$ is a group.
For your question, just take $g = -1$.
