Let $\mathbb Q^{\times}/\mathbb Q^{\times^2}$ denote the group of rational numbers modulo squares. This means that we regard two nonzero rational numbers $x_1, x_2$ as equivalent if the ratio $x_1/x_2$ is the square of a rational number.
Consider the definitions in the following theorem -
Also the following proposition and corollary-
Now it is written that Since $\Gamma'$ is finite, but I can't see that how, can anyone explain with simple example of elements of $\Gamma'$? Thanks.
OK, here is an answer with an example. Let us consider the sample elliptic curve with affine equation over the rationals $$ y^2 = x(x-10)(x-17)\ , $$ which has $e_1,e_2,e_3$ equal to $0,10,17$ (in this order, say).
Here are some rational points on the curve, and their maps in $\Bbb Q^\times/(\Bbb Q^\times)^2$. For psychological reasons, a class $\bar a$ of some rational number $a$ in this group (with the inherited multiplication) will be written below as "$a$ modulo squares". Alternatively one can spell $\bar a$ as "$a$ modulo square factors", i.e. the operation we are allowed to do is to get rid of squares in the factorization of $a\in\Bbb Q$, $a\ne 0$. As warming up, we have for instance in this group for the first some random numbers i am choosing: $$ \begin{aligned} 20 &= 2^2\cdot 5=5&&\text{modulo squares}\\ 24 &= 2^3\cdot 3=2\cdot 3&&\text{modulo squares}\\ -4 &=-1&&\text{modulo squares}\\ \frac 23&=2\cdot 3&&\text{modulo squares}\\ 1024 &=2^{10}=1&&\text{modulo squares}\\ \end{aligned} $$ and so on. I think that now the question is answered, but let us go till the end.
The computer is generating some rational points on the given curve. For each point $P$ in the list we compute its image $\delta(P)$ via $\delta$. This image is a triple, and each of its components is (well defined only after it is) taken "modulo squares".
For instance: $$ \begin{aligned} T&=(0,0) \\ \delta(T) &=((0-10)(0-17), \ 0-10, \ 0-17)\\ &=(170, -10, 17)&&\text{mod squares,} \\[2mm] P&=(1,12)\\ \delta(P) &=(1-0,\ 1-10,\ 1-17))\\ &=(1, -9, -16)=(1,-1,-1)&&\text{mod squares,} \\[2mm] Q&=(8,12)\\ \delta(Q) &=(8-0, \ 8-10,\ 8-17))\\ &=(8, -2, -9)=(2,-2,-1)&&\text{mod squares,} \\[2mm] R&=(45, 210)\\ \delta(R) &=(45-0, \ 45-10,\ 45-17))\\ &=(45, 35, 28)=(5,35,7)&&\text{mod squares,} \\[2mm] S&=(10125, 1017450)\\ \delta(S) &=(10125-0, \ 10125-10,\ 10125-17))\\ &=(3^4\cdot 5^3, \ 5\cdot 7\cdot17^2, \ 2^2\cdot7\cdot19^2)\\ &=(5,35,7)&&\text{mod squares,}\\ \\[2mm] X&=(a,b)\ ,\text{ with}\\ a&=\frac{75414992906898}{313242821761}\ ,\\ b &=\frac{618120625766999640108}{175316055726018241}\ ,\\ \delta(X) &=(a-0, \ a-10,\ a-17))\\ a-0&=\frac{2\cdot 3^2\cdot 2046881^2}{359^2 \cdot 1559^2}=2 &&\text{mod squares,}\\ a-10&=\frac{2\cdot 3005881^2}{359^2 \cdot 1559^2 }=2 &&\text{mod squares,}\\ a-17&=\frac{47^2\cdot 178127^2}{359^2 \cdot 1559^2 }=1 &&\text{mod squares,}\\ \delta(X)&=(2, 2, 1) &&\text{mod squares.} \end{aligned} $$ It is maybe good to note that in all cases we have a product of the components in the image triple which is one modulo squares.
Exercise: Compute $\delta(T)$ for the other two torsion points $T=(10,0)$ and $T=(17,0)$.
The observation is now that for all values from above of a possible $x_0\in\Bbb Q$ (such that there it can be completed, lifted with an $y_0$...) when we compute $(x_0-e_1)=(x_0-0)$, $(x_0-e_2)=(x_0-10)$, $(x_0-e_3)=(x_0-17)$, and eliminate all square factors from their factorizations, then the remained primes are among the factors of $\Delta=(0-10)(0-17)(10-17)$. And indeed, such prime factors like $3005881$ in $a-10$ from the last example are constrained to appear to even power. In all computations of a $\delta$-image, we have obtained a number of the shape $$ (-1)^s2^t5^u7^v17^w $$ with powers $s,t,u,v,w$ among zero or one.
Now the short answer to the question:
The group $\Gamma'$ from the cited result in the OP can thus be seen as, identified as (or injectively mapped to) the group of all $2^5$ numbers of the above form, with multiplication modulo squares as operation. It is a finite group.
In general, we compute $\Delta$, consider its prime factors, count them, if they are $f$, then $\Gamma'$ has $2^{f+1}$ elements, the $+1$ in the power considering the non-square rational unit $(-1)$. The operation is multiplication modulo squares. If we pass to an additive writing, than this group is isomorphic to $(\Bbb Z/2)^{f+1}$.