Max number of configurations of a Turing Machine
Yes, that's right.  If there are $k$ possible configurations, then any such Turing machine either runs in time at most $k$, or it loops forever.  That's because if it runs for at least $k+1$ time steps, some configuration must be repeated; and if a configuration is repeated, it will continue repeating forever.

Let $|\Gamma|=2^c$, i.e., $c = \lg |\Gamma|$.  Then $|\Gamma|^{f(n)} = (2^c)^{f(n)} = 2^{c f(n)} = 2^{O(f(n)}$.  For similar reasons, $\vert Q \vert \cdot f(n) \cdot \vert \Gamma \vert ^{f(n)} = 2^{O(f(n)}$.  So, the number of possible configurations of a $f(n)$-space-bounded Turing machine is $2^{O(f(n)}$, and thus any such machine must either halt in time at most $2^{O(f(n)}$ or loop forever.