Why is it that $\mathscr{F} \ne 2^{\Omega}$?

The $\sigma$-algebra generated by the events $\{\omega \in \Omega: \omega_n = W \}$ is the so-called Borel $\sigma$-algebra on $\Omega = \{H,T\}^\mathbb{N}$.

One can show, by transfinite induction (so you need some set-theory background) that there are at most $|\mathbb{R}| = 2^{\aleph_0}$ many Borel sets, while the power set of $\Omega$ has $2^{|\Omega|} = 2^{|\mathbb{R}|}$ many subsets, which is more by Cantor's theorem. So $\mathcal{F}$ is much smaller than the power set of all subsets of $\Omega$. But you only need the sets in $\mathcal{F}$. One can also use the Axiom of Choice to find non-Borel sets (the characteristic function of an ultrafilter, e.g.).

Filling in all the details requires some theory the author presumably did not want to assume the reader to know about.

Blackwell & Diaconis ["A non-measurable tail set", in Statistics, probability and game theory, pp. 1–5, IMS Lecture Notes Monograph Series, vol. 30, 1996] give an example of a subset of $2^{\Bbb N}$ that is not an element of $\mathscr F$. Their construction uses a free ultrafilter $\mathscr U$ on $\Bbb N$. Let $E\subset 2^{\Bbb N}$ consist of those $a=(a_1,a_2,\ldots)\in 2^{\Bbb N}$ (thus each $a_i$ is either $0$ or $1$) such that $N_a:=\{i\in\Bbb N:a_i=1\}\in\mathscr U$. Then $E\notin\mathscr F$.