Is the composite of two simple radical extensions another simple radical extension?

I don't think they ever state that composition of simple radical extension is also a radical extension. Moreover, I simply do not think it's true. For example, I don't think that $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ is a simple radical extension: its Galois group is not cyclic.

The only thing that they claim is that the composition of root extensions is also a root extensions, and that's easy. Suppose we have two root extensions: $F = F_0 \subset F_1 \subset \ldots \subset F_p$, and $F = G_0 \subset G_1 \subset \ldots \subset G_q$, such that $F_{i+1} = F_i(\sqrt[n_i]{a_i})$ for some $a_i \in F_i$, and same for $G_i$. WLOG we can assume that $p = q$.

Then, the composition of $F_p$ and $G_p$ is also a root extension: since $G_1 = G_0(\sqrt[k_0]{b_0})$, $b_0 \in G_0 = F = F_0$. Thus, we set $H_1 = F_1(\sqrt[k_0]{b_0}) = F_0(\sqrt[n_0]{a_0})(\sqrt[k_0]{b_0})$, and we see that $H_1$ is a root extension, and $F_1 \subset H_1$, $G_1 \subset H_1$ -- in fact, $H_1$ is the composition of $F_1$ and $G_1$. Similarly, since $a_1, b_1 \in H_1$, we set $H_2 = H_1(\sqrt[n_1]{a_1})(\sqrt[k_1]{b_1})$, and again $H_2$ is a root extension, and the composition of $F_2$ and $G_2$. Continuing this way, we obtain that $H_p$ is a root extension and the composition of $F_p$ and $G_p$.