Suppose $\text{Im}(f(z)) < M$ for some analytic $f: \Omega - \{a\} \rightarrow \mathbb{C}$. Then why does the Möbius transformation
$$ g(z) \mapsto {iMf(z) \over 2iM - f(z)} $$
map $z$ into the interior of the disc $|z| < M$?
Is this due to us discovering the bound on $|g(z)|$ due to an inequality calculation, or is there some major theorem concerning Möbius transformations which immediately yields this result?
That depends on the notion of "immediately".
The major theorem is that Möbius transformations map circles and lines to circles and lines - from the viewpoint of the Riemann sphere, they map circles to circles; a straight line in $\mathbb{C}$ is just a circle through $\infty$.
And Möbius transformations are conformal mappings, i.e. they preserve angles.
Then you just need to see what curve the line $L = \{ w : \operatorname{Im} w = M\}$ is mapped to by the Möbius transformation $T\colon w \mapsto \frac{iMw}{2iM-w}$. The point $\infty$ is mapped to $-iM$, and the point $iM$ is mapped to $iM$. Since $L$ is perpendicular to the imaginary axis, and the imaginary axis is mapped to itself by $T$, the circle $T(L)$ intersects the imaginary axis in a right angle. That means the centre of the circle lies on the imaginary axis. Since $T(L)$ intersects the imaginary axis in $iM$ and $-iM$, it must hence be the circle $C = \{ \zeta : \lvert\zeta\rvert = \lvert M\rvert \}$.
It remains to see whether the half-plane $\operatorname{Im} w < M$ is mapped to the inside or to the outside of $T(L)$. Since $T(0) = 0$, it is mapped to the inside of the circle if $M > 0$ and to the outside if $M < 0$.