The splitting field of $x^4 - 2x^2 - 6$ is $\mathbb{Q}(\alpha, \sqrt{-6}) = \mathbb{Q}(\alpha, \beta)$ where $\alpha = \sqrt{1+\sqrt{7}}$, $\beta = \sqrt{1-\sqrt{7}}$. From the first representation, $x^4 - 2x^2 - 6$ being irreducible (say, Eisenstein for $2$) and $\sqrt{-6}$ being non-real it follows that the extension (and hence the Galois group) has order $8$.
Now one can sneakily show that the group is $D_8$ as follows -- it is not Abelian, as then by FTGT any intermediate extension must be Galois, whereas $\mathbb{Q}(\alpha) : \mathbb{Q}$ is not. On the other hand, $(\alpha \to -\alpha)$ (and fix the rest) and $(\sqrt{-6} \to -\sqrt{-6})$ (and fix the rest) are two distinct morphisms of order $2$, hence the group cannot be quaternion. So it must be $D_8$.
But what about a set of generators? I think that $(\sqrt{-6} \to -\sqrt{-6})$ can be the reflection one but I can't think of a suitable rotation one.
Any help appreciated!
The roots of $X^4-2X^2-6$ are $\alpha_1=\sqrt{1+\sqrt{7}},\alpha_2=\sqrt{1-\sqrt{7}}$, $\alpha_3=-\alpha_1$ and $\alpha_4=-\alpha_2$.
You have already determined that the Galois group is isomorphic to the dihedral group $D_4=\langle\rho,\sigma\mid \rho^4=1,\sigma^2=1,\sigma\rho\sigma^{-1}=\rho^{-1}\rangle$.
To give explicit generators, we seek an element $\rho$ of order $4$ and and element $\sigma$ of order $2$ that fulfill the relation $\sigma\rho\sigma^{-1}=\rho^{-1}.$
An element $f$ of the Galois group can send $\alpha_1$ to any of the four other roots, then there are two possibilites left for $f(\alpha_2)$. (Since $f(\alpha_1)$ and $f(\alpha_3)=f(-\alpha_1)=-f(\alpha_1)$ are already determined.)
Take $\rho:\begin{cases}\alpha_1 \longmapsto \alpha_2 \\ \alpha_2\longmapsto -\alpha_1\end{cases}$ and $\sigma:\begin{cases}\alpha_1 \longmapsto \alpha_2 \\ \alpha_2\longmapsto \alpha_1\end{cases}$. Verify as an exercise that these do the job.