A Möbius transformation maps circles and lines to circles and lines. What exactly does that mean? The title pretty much says it all. I am also looking for a concrete example if possible. I have looked at the proof, but I'm not exactly sure what it means because I am kind of confused on what the theorem actually says. Does it mean that if g(z) is a circle and f(z) is a Möbius transformation then f(g(z)) is a circle as well? 
 A: The meaning is that if $C\subseteq\mathbb{C}$ is a circle or line, and $\varphi$ a Möbius transform, then indeed $\varphi[C]$ would be either a line or a circle. However, the image of a circle/line might be a line or a circle, regardless of which of these $C$ is.
Considering the homeomorphism between the complex plane and the Riemann sphere, every line/circle is a circle on the sphere (lines are simply circles 'through infinity'). This observation means that Möbius transforms map circles in the Riemann sphere to circles.
A: No. It means if $C$ is a line ("circle with infinite radius centered in $\infty$") or a circle, then $f(C)$ is also a line or a circle. It does neither imply $f(\text{line}) = \text{line}$ nor $f(\text{circle}) = \text{circle}$.

For reference, the cayley-transform maps $\mathbb{R}$ to $S^1$ and its inverse maps $S^1$ to $\mathbb{R}$, which is a line and a circle respectively.
A: Consider the image of a line in $\mathbb{C}$ projected onto the Riemann sphere. As you follow the line in one direction out towards infinity, its projection onto the sphere tends to the north pole.
As you follow the line in the opposite direction, its projection also tends to the north pole!

If you take any line or circle in $\mathbb{C}$, and transform it using a Möbius transformation, then the resulting set will also be a line or a circle in $\mathbb{C}$.
All we really need to consider is the case $w = \frac{1}{z}$. Immediately, we should expect a problem at $z=0$.
Let $S$ be a circle of radius $r$ centered at $\alpha$. We know that $(z-\alpha)(\overline{z}-\overline{\alpha}) = r^2$ (this is just the equation for a circle), so
$$ z\overline{z}-\alpha\overline{z}-\overline{\alpha}{z} = r^2-|\alpha|^2 \\
\frac{1}{w\overline{w}}-\frac{\alpha}{\overline{w}}-\frac{\overline{\alpha}}{w} = r^2-|\alpha|^2
$$
If there is no point in $S$ that goes through the origin, then the resulting image is going to just be a circle!
But if a point in $S$ does go through the origin, then $r = |\alpha|$, so $1-\alpha w - \overline{\alpha}\overline{w} = 0$.
This means that $\textrm{Re}(\alpha w) = \frac12$. Solving for this, we find that $\textrm{Re}(w)\textrm{Re}(\alpha)-\textrm{Im}(w)\textrm{Im}(\alpha) = \frac12$, which we should recognize as the parametric form of a line.
Therefore, $f(z) = \frac{1}{z}$ maps a circle to a circle, unless that circle goes through the origin, in which case it becomes a line.
Then, $F(z) = \frac{az+b}{cz+d}$ is just the composition of $\frac{1}{z}$ on the left and right by linear functions, which are just scalings and translations!
