Showing $K(\sqrt \alpha)/F$ is Galois if and only if $\sigma(\alpha)/\alpha$ is a unit and a square.

As I mentioned in the comments, the problem is that in general $\sqrt{\sigma(a)}$ might not be in $K(\sqrt{a})$. Notice that if $E$ is the Galois closure of $|K(\sqrt{a}):F|$ then $\sqrt{\sigma(a)}\in E$ so $|K(\sqrt{a}):F|$ is Galois if and only if $E=K(\sqrt{a}) \iff \sqrt{\sigma(a)}\in K(\sqrt{a}), \forall\sigma\in G$
Suppose $a$ is not a square.
We only need to show that $\sqrt{\sigma(a)}\in K(\sqrt{a}) \iff \sqrt{\sigma(a)}=k\sqrt{a},k\in K$. One direction is immediate: $\sigma(a)=k^2a\implies \sqrt{\sigma(a)}=k\sqrt{a}\in K(\sqrt{a})$.
For the other direction: $\sqrt{\sigma(a)}\in K(\sqrt{a}) \implies \sqrt{\sigma(a)}=k_1+k_2\sqrt{a}$ with $k_{1,2}\in K(a) \implies \sigma(a)={k_1}^2+a{k_2}^2+2k_1k_2\sqrt{a}$. Since $\sqrt{a}\not\in K(a)$ we must have $k_1=0$ or $k_2=0$, the later implies $\sqrt{\sigma(a)}\in K \implies\sqrt{a}\in K$ yielding a contradiction. Thus we must have $k_1=0$ and $\sigma(a)={k_2}^2a$ with $k_2\in K$.

The question actually extends to the following setting : $F$ contains the group $\mu_n$ of $n$-th roots of $1$, char$F$ does not divide $n$, $K/F$ is galois with group $G$; then for $a\in K^\times$, $L:=K(\sqrt [n]a)$ is galois over $F$ iff $s(a)/a \in $ for all $s\in G$.

Proof: For clarity, recall that the (ambiguous) notation $\sqrt [n]a$ just means an arbitrarily chosen $n$-th root of $a$ in a separable closure of $F$. Moreover, the extension $L/K$ depends only on the class $[a]$ of $a$ mod ${K^\times}^n$, so it will be convenient to write $L=K(\sqrt [n]{[a]})$. The advantage of this new notation is that $K(\sqrt [n]{[a]})=K(\sqrt [n]{[b]})$ iff $[a]=[b]$ in ${K^\times}/{K^\times}^n$.

The separability of $L/F$ being ensured by the hypothesis on char$F$, we must only show normality. Any $s\in G$ can be extended to an $F$-homomorphism $\bar s$ ("embedding") of $L$ into a separable closure. The normality of $L/F$ is then equivalent to the stability of $L$ under every $\bar s$. By definition, $(\bar s(\sqrt [n]a))^n=\bar s(a)=s(a)$, so the preliminary remarks above show that the sought for normality is equivalent to $[s(a)]=[a]$, in other words $s(a)/a\in {K^\times}^n$ for all $s\in G$. Note that all along, we only used the multiplicative structure of the fields involved, not their additive structure. This approach also allows (but cohomological tools are required) to describe explicitly the group $\bar G=Gal(L/K)$ starting from $G$. For example, if $n=2$ and $G=C_2 \times C_2$, one can derive criteria for $\bar G$ to be $C_2\times C_2 \times C_2$, or $D_4$, or $H_4$.