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.
Since $K=\mathbb{Q}(\zeta,\sqrt[5]{5})$, an element $\sigma\in\mathrm{Gal}(K/\mathbb{Q})$ is completely determined by what $\sigma(\zeta)$ and $\sigma(\sqrt[5]{5})$ are. Note that any $\sigma\in\mathrm{Gal}(K/\mathbb{Q})$ must have
$$\sigma(\zeta)=\zeta^k\text{ for some }1\leq k\leq 4,\qquad\quad \sigma(\sqrt[5]{5})=\zeta^r\sqrt[5]{5}\text{ for some }0\leq r\leq 4$$
(These comprise the $20$ different elements of $\mathrm{Gal}(K/\mathbb{Q})$.)
To understand $\mathrm{Gal}(K/\mathbb{Q})$, try to write it using a presentation (generators and relations). Can you think of any particular elements of $\mathrm{Gal}(K/\mathbb{Q})$ that, taken together, generate the entire group? Try to treat the two generators of the field "orthogonally" - find one automorphism that permutes the various powers of $\zeta$ and ignores $\sqrt[5]{5}$, and another automorphism that permutes the various 5th roots of $5$ and ignores any $\zeta$s (or at least, any $\zeta$s that aren't part of one of the 5th roots of $5$). Then combinations of these two should allow you to produce any desired effect on both $\zeta$ and $\sqrt[5]{5}$.
Then, determine the relations between those generating elements. That should make it easier to figure out the subgroups of $\mathrm{Gal}(K/\mathbb{Q})$ (and hence, the subfields of $K$).
Best Answer
The important fact you need to use is the following:
Theorem Let $\sigma : L \to F$ be a morphism of fields and fix algebraic closures of $\overline{L}$ and $\overline{F}$ of $L$ and $F$, respectively. Let $a \in \overline{L}$ with minimal polynomial $p \in L[x]$. Let $S = \{f : L(a) \to \overline{F} \mid f|_L = \sigma\}$ and let $T$ be the set of roots of $\sigma(p) \in F[x]$ in $M$. Then the map $$f \mapsto f(a) : S \to T$$ is a bijection.
You can find an exposition of this and related results in Dr. Romyar Sharifi's excellent abstract algebra notes (roughly around section 6.7).
In your case, you want to know that there is an automorphism of $K$ over $\mathbb{Q}$ which sends any given root of your polynomial to another. In other words, you want to extend the inclusion $\mathbb{Q} \to K$ to a morphism $K \to K$ (which will neccesarily be an automorphism since $K / \mathbb{Q}$ is finite) such that this extension maps one root to another. This is possible by the above theorem! First, choose a root $r_1$ which we want to send to a root $r_2$. The theorem says there is a unique extension of the inclusion $\mathbb{Q} \to K$ to a map $\mathbb{Q}(r_1) \to K$ such that $r_1 \mapsto r_2$. If $\mathbb{Q}(r_1) = K$ then we're done, otherwise choose some root $r_3 \in K \setminus \mathbb{Q}(r_1)$ and extend to $\mathbb{Q}(r_1)(r_3) = \mathbb{Q}(r_1, r_3) \to K$ by sending $r_3 \mapsto r_3$. Eventually we will adjoin enough roots to get our desired extension $K \to K$.
Hope this helps! Let me know if you have any questions about the above.