I somehow don't really understand how to choose the points for a moebius transform.
I know that a moebius transform maps circles and lines to circles and lines and that it is a conformal(biholomorphic) map.
Why do we always choose the three points on the boundary of the corresponding domain?
When we for example want to construct a moebius transform from the open unit disk to the upper half plane, then we choose three points on the boundary of the unit disk which we want to map to the boundary of the upper half plane which is the real axis. Why is it necessary that all these points are on the corresponding boundary?
How are the points exactly mapped to each other?
It seems that one cannot just map the 3 points in any order to the image points. I have read that one has to keep the orientation in mind. What is meant by that?
Do I have also to keep something in mind when I am mapping the 3 image and pre-image points to $0,1,\infty$?
How are only 3 point-pairs with information on orientation enough to specify the mapping
For me this seems unintuitive. I could just change the target domain a little bit so that the boundary points I chose still lie on the boundary of the domain. Wouldn't I get the exact same moebius transform although my target domain is different now?
Generalized circles (circles and lines) are determined by three points. If your Mobius transformation maps three points from one generalized circle to another, then it is automatically guaranteed to map the first generalized circle to the second. The converse is also true: if a Mobius transformation maps one domain to another ("domain" in this context meaning a disk or abscissa aka half-plane) then it must map the first boundary to the second (this is simply a topological fact about continuous maps), meaning every point on the first boundary is mapped to a point on the second boundary.
Suppose $a,b,c$ are located in order counterclockwise around a closed disk. If $f$ is a Mobius transformation that maps this to a second closed disk, and in particular maps $a,b,c$ to $x,y,z$ (respectively), then $x,y,z$ are also located counterclockwise around the second closed disk.
Thus, if you want to construct a Mobius transformation $f$ that maps one closed disk to another, you can pick three points $a,b,c$ around the first boundary and $x,y,z$ around the second and construct $f$ to map $a,b,c$ to $x,y,z$ respectively, but you need to make sure $a,b,c$ and $x,y,z$ are located around their domains either both counterclockwise or both clockwise for this to work. If you don't do this, $f$ will map the first domain to the complement of the second domain (i.e. it will swap it inside out).
This applies when using $0,1,\infty$ too. (Remember also, from the perspective of the Riemann sphere, lines in the plane are just circles and abscissas are just disks. That applies to the real number line which bounds the upper and lower half-planes.)
You can't change the domain without changing the three points on the boundary, because as mentioned, the domains we're talking about are closed disks (from the perspective of the Riemann sphere), which are determined by three boundary points and an orientation. (If you adopt the convention that all boundaries should be oriented counterclockwise, for example, then the orientation of the three points - i.e. a choice of cyclic ordering - will determine which side of the boundary the domain is actually on.)