As discussed in the comments, it is usually easiest to piece something like this together by hacking around than to follow a methodical approach. But I thought you might be interested in seeing a methodical approach written up.
In general, let's understand the Galois group $G$ of the splitting field of $x^4+bx^2+c$. Let the roots of $x^4+b x^2+c$ be $\pm \alpha$ and $\pm \beta$. We will assume that the polynomial doesn't have repeated roots. This is equivalent to $(b^2-4c)c \neq 0$.
Any Galois symmetry must either take the pair $\{ \alpha, -\alpha \}$ to itself, or to $\{ - \beta, \beta \}$, because these are the two two-element subsets of the roots which sum to $0$. So the group is a subgroup of the dihedral group $D_8$. I like to think of $D_8$ as the symmetries of a square, with $\pm \alpha$ and $\pm \beta$ at diagonally opposite corners of the square.
You act as if there are two four element subgroups of $D_8$, but there are really three: $C_4$, the copy of $V_4$ generated by reflections over lines parallel to the sides of the square and the copy of $V_4$ generated by reflections over the diagonals of the square. The last $V_4$ doesn't act transitively, so you rule it out at an earlier stage, but I'd rather keep it around.
Reflections over lines parallel to the sides of the square: Consider the element $\gamma: =\alpha \beta$ in the splitting field. If the full $D_8$ acts on the roots, then the orbit of $\gamma$ is $\pm \gamma$ and the stabilizer of $\gamma$ is this $V_4$. In general, if the group is $G$, then the stabilizer of $\alpha \beta$ is $G \cap V_4$. So $G$ is contained in this $V_4$ if and only if $\alpha \beta$ is fixed by the full Galois action, if and only if $\alpha \beta$ is rational.
Now, $(\alpha \beta)^2 = c$. So we get that $G$ is contained in this $V_4$ if and only if $c$ is square.
Reflections over the diagonals of the square: The element $\alpha^2$ is stabilized by this copy of $V_4$; so is the element $\beta^2$. So $G$ is contained in this $V_4$ if and only if $\alpha^2$ and $\beta^2$ are rational. Now, $\alpha^2$ and $\beta^2$ are the roots of $x^2+bx+c$, and the roots of this quadratic are rational if and only if $b^2-4c$ is square.
So $G$ is contained in this $V_4$ if and only if $b^2-4c$ is a square.
The group $C_4$: Again, I think of $C_4$ as a subgroup of the symmetries of a square -- specifically, the rotational symmetries. I am gong to find an element $\delta$ whose stabilizer is $C_4$; this will play a role analogous to $\gamma$ in the first section and $\alpha^2$ in the second.
I found $\gamma$ and $\alpha^2$ just by guessing, but $\delta$ took me a little thought.
I'd like a polynomial in $\alpha$ and $\beta$ which has odd degree in each, so that it is not fixed under reflection over any of the diagonals of the squares. We saw above that $\alpha \beta$ doesn't work -- it's stabilizer is $V_4$. Let's try a linear combination of $\alpha \beta^3$ and $\alpha^3 \beta$. A $90^{\circ}$ rotation of the square takes $\alpha \mapsto \beta$ and $\beta \mapsto - \alpha$, so it negates and switches the preceding monomials. In short, we take
$$\delta = \alpha \beta^3 - \alpha^3 \beta.$$
If all of $D_8$ acts, then the orbit of $\delta$ is $\pm \delta$. So, as above, the Galois group is contained in $C_4$ if and only if $\delta$ is rational.
Now,
$$\delta^2 = (\alpha^2 \beta^2) (\alpha^2 - \beta^2)^2 = c \cdot (b^2-4c).$$
(Remember that $\alpha^2$ and $\beta^2$ are the roots of $x^4+bx^2+c$.)
So $G \subseteq C_4$ if and only if $c(b^2-4c)$ is square.
In your case, we have $c \cdot (b^2 - 4c) = 2 \cdot (4^2-4 \cdot 2) = 16$, so your Galois group is contained in $C_4$.
By the way, notice that the intersection of any two of these groups is contained in the third. Correspondingly, if any two of $c$, $b^2-4c$ and $c \cdot (b^2-4c)$ are square, so is the third.
Best Answer
The field $K$ is an extension of $F_1$ of degree $8$. The minimal polynomial of $\sqrt[8]2$ over $F_1$ divides $x^8-2$ since $\sqrt[8]2$ is a root. Since adding $\sqrt[8]2$ to $F_1$ generates an extension of degree $8$ of $F_1$, the minimal polynomial of $\sqrt[8]2$ over $F_1$ must have degree $8$, hence it is $x^8-2$. Therefore the extension is cyclic, hence its Galois group is cyclic.
To check this last claim, use Proposition $36$ of chapter 14 section 7 : since $L = \Bbb Q(\sqrt 2, i)$ contains the $8^{\text{th}}$ roots of unity, the extension $L(\sqrt[8]2)/L = K/L$ is cyclic. Therefore, $\mathrm{Gal}(K/L)$ is a cyclic subgroup of $\mathrm{Gal}(K/F)$. Since $\mathrm{Gal}(K/L)$ has index $2$ in $\mathrm{Gal}(K/F)$, it is normal, so let $\sigma_1 \in \mathrm{Gal}(K/F) - \mathrm{Gal}(K/L)$. Check that $\sigma_1$ generates $\mathrm{Gal}(K/F)$ using the above information.
EDIT : I admit I can't finish the proof... I thought it had a neat end. I'll leave it there just in case someone can finish it.
Perhaps using chapter 14.7 is a little harsh because it comes later on, but if you look at the proof of that proposition you see that the ideas are not quite fancy, you can read them already and get a good idea of what it is about.
Hope that helps,