How does O transform this sum like that?
Intuitively, the $O$ can swallow any constant factor. Here, we do so because the sum we get is on the cheat sheet and we can proceed more easily.

Formally, we need to apply the definition of $O$, once in either direction. I do not think the first term makes a whole lot of sense. What the authors probably mean is

$\qquad\displaystyle \sum_{h=0}^{\lg n} \frac{n}{2^{h+1}} \cdot f(h)$

with some $f \in O(h)$. Then it remains to note that

$\qquad\displaystyle \sum_{h=0}^{\lg n} \frac{n}{2^{h+1}} \cdot f(h) 
  \quad\leq\quad \sum_{h=0}^{\lg n} \frac{n}{2^{h+1}} \cdot c \cdot h
  \quad=\quad \frac{c}{2} \cdot n \cdot \sum_{h=0}^{\lg n} \frac{h}{2^{h}}$

for all $n \geq n_0$, with $c,n_0$ the constants from $f \in O(h)$ (unfold the definition).
Applying the definition again concludes the proof.

Note that if we had $f \in \Theta(h)$, we would get a similar lower bound and could thus show a $\Theta$-bound for the sum.