If $E/k$ is solvable, $F$ a field containing $k$ and $K$ is Galois solvable over $K$ with $E\subset K$ then $KF$ is Galois over F, why ?
Nothing is said about $F$, it is just an arbitrary field extension of $k$. It occurs in the middle of the following proposition here The mentioned theorem $1.12$ proves just the subgroup property not that it is Galois
This is a general fact:
Proof: $K/k$ is finite separable, hence simple, i.e. $K=k[\alpha]$ for some $\alpha$ algebraic (and separable over $k$). Thus, it is easy to see that $FK=F[\alpha]$ (on one hand, $\alpha \in K \subseteq FK,$ so $F[\alpha] \subseteq FK$; on the other hand, $F[\alpha]$ contains $F$ and $k[\alpha]=K$, hence $F[\alpha] \supseteq FK$). Now the minimal polynomial $m_{\alpha, F}$ clearly divides $m_{\alpha, k},$ which shows that $\alpha$ is separabe over $F$. Also, all the roots of $m_{\alpha, k}$ are in $K$ (since $K/k$ is normal), hence all the roots of $m_{\alpha, F}$ are contained in $FK$ (again, because $m_{\alpha, F} \mid m_{\alpha, k}$, and $K \subseteq FK$). This shows that $FK/F$ is normal as well. Altogether, $FK/F$ is a Galois extension (and as a bonus, it is finite and $[FK:F]$ divides $[K:k]$).
Remark: Since general Galois extensions are just unions of finite Galois ones, I guess the statement hold even without the finiteness assumption.