I want to find a linear fractional transformation which maps the region $D$ of the $z$-plane onto the region $G$ of the $w$-plane, where
$D=\{z;|z-1|<1\},~G=\{w;\Re w>0\}$
This is an exercise in my complex variable book and I can solve it by the next section which is about the Mapping of the upper half-plane onto the unit disk as follows
First we can map $G$ onto $D$ by $w=\frac{iz-i}{iz+i}+1=\frac{2iz}{iz+i}$, so we have $z=\frac{iw}{2i-iw}=\frac w{2-w}$.
Thus we conclude that $w=\frac{z}{2-z}$ maps $D$ onto $G$.
But I want to know what is the direct way in order to map $D$ onto $G$. I've seen some other books, but they've used the cross ratio, but my book hasn't said anything about the cross ratio. Could anyone help me, please?
This is the results I've found now. Let $w=\frac{az+b}{cz+d}$, by the following conditions
$$w(0)=0,~w(2)=\infty$$ we will obtain the map $w=\frac z{2-z}$. But I don't know why these conditions gives the correct map?
Since you want to send an open disk to an open half plane, you want to send the boundary, which is the circle $|z-1|=1$ to the boundary, which is $\Re z=0$. They share a point, actually $z=0$, so it is natural to send $0$ to $0$. The opposite point to $0$ in the circle is $z=2$, and we will send it to $\infty$. Now, the centre of the circle, $z=1$, can stay fix, to be sure you send the interior of the circle to the right half plane (if you send $1$ to $-1$, then you send the disk to the opposite half plane). So you want $$w(0)=0, \ w(2)=\infty \mbox{ and } w(1)=1.$$ This three conditions give you the map you want.
Observe, however, that you can take $w+ia$ for any $a\in \mathbb R$ and the function also verifies what you want. Or even $1/w$. Or the composition with any linear fractional transformation that sends the half plane to the half plane.
EDIT: Another (may be better) way to construct such maps (from a disk to a disk or half-plane) is to send three points of the boundary (=circle) to three points on the boundary (circle or line): this will send the interior of the disk to either the interior or the exterior of the circle (or one of the half-planes, in the case of lines); changing the order of two of the points if necessary will give you the right map.
For example, in your case you can send $w(0)=0$, $w(2)=\infty$ and $w(i+1)=-i$. So $w(z)=-z/(2-z)$ and then $w(1)=-1$, which is not correct. But if you instead send $w(2)=-i$ and $w(i+1)=\infty$, so $w(z)=-(i+1)z/(2z-2(i+1))$, then $w(1)=(1-i)/2$ and it works. Of course it will better to send $w(0)=0$, $w(2)=\infty$ and $w(i+1)=i$, which gives your original map, but it was just to give you a different example.