Galois Group of $x^p - 2$, $p$ an odd prime

Here it might be best to realize the Galois group as a group of permutations of the roots of the polynomial of interest (as in user64494's comment under the OP). You have already observed that the roots of $x^p-2$ are $x_i=\zeta_p^{i-1}\root p\of 2, i=1,2,\ldots,p.$ You also know that the splitting field is of degree $p(p-1)$, so that is also the order of the Galois group.

Let us consider the action of your automorphism $\tau$ defined by $\tau(\zeta_p)=\zeta_p$ and $\tau(\root p\of 2)=\zeta_p\root p\of 2$. So we see that $\tau(x_i)=x_{i+1}$, if $i<p$, and $\tau(x_p)=x_1$. The action of $\tau$ on the chosen indexing of the roots thus corresponds to the $p$-cycle $\tau=(123\cdots p).$

On the other hand the automorphism $\sigma_a:x_1\mapsto x_1, \zeta_p\mapsto \zeta_p^a,1\le a<p,$ keeps the real root $x_1$ fixed, and permutes the others according to the rule $x_i=x_1\zeta_p^{i-1}\mapsto x_1\zeta_p^{a(i-1)}=x_{1+a(i-1)}$, where the subscript is calculated modulo $p$. You see that all these share $x_1$ as a fixed point (this was also clear from your construction of $\sigma$:s as elements of the Galois group $G_2=Gal(\mathbb{Q}(x_1,\zeta_p)/\mathbb{Q}(\zeta_p))$.

You can either look at all these automorphisms as elements of $S_p$. This works beautifully, once you have found a generator of $G_2$. This is equivalent to finding a generator of the multiplicative group $\mathbb{Z}_p^*$, i.e. a primitive root. There is no general formula for such a generator, so I won't say much about that (this may be a cause of your difficulties). We simply know that one exists! But if $p$ is fixed, say $p=5$ or another smallish prime, then I recommend this way, as you can easily calculate with permutations.

PVAL is strongly hinting at the possibility that you may get a semi-direct product of $G_1$ and $G_2$. Indeed, you will see that one of the two subgroups is stable under conjugation by elements of the other. Which way does it work? I'm a mean dude and won't tell you! But you do remember that the fixed field of a normal subgroup is itself Galois over the base field. So which of the fields $\mathbb{Q}(\zeta_p)$ or $\mathbb{Q}(\root p\of2)$ is Galois over the rationals? The group of automorphisms associated to that field should be a normal subgroup of the big Galois group. After figuring that out, you can start studying the effect of either $\sigma_a\tau\sigma_a^{-1}$ or $\tau\sigma_a\tau^{-1}$ all according to which feels more interesting...

I thought I'd add a realization of this Galois group that can be analysed easily with direct computation, since it turns out that the Galois group is actually quite easy to describe.

Let $G = Gal(X^p-2)$ (or indeed $Gal(X^p-q)$ for any prime $q$). Then, as observed, the splitting field is $L = \mathbb{Q}(\zeta,\sqrt[p]{2})$. So any element $\varphi \in G$ is determined by $\varphi(\zeta)$ and $\varphi(\sqrt[p]{2})$. But $\varphi(\zeta) = \zeta^a$ for some $1 \leq i \leq p-1$ and $\varphi(\sqrt[p]{2}) = \zeta^b \sqrt[p]{2}$ for some $0 \leq b \leq p-1$ We thus get a map:

$$\chi:G \rightarrow G' :=\bigg\{\begin{pmatrix} a & b \\ 0 & 1\\ \end{pmatrix} : a \in \mathbb{F}_p^{\times}, b \in \mathbb{F}_p\bigg\} \subset \mathrm{GL}(2, \mathbb{F}_p) $$

$$\chi(\varphi)= \begin{pmatrix} a & b \\ 0 & 1\\ \end{pmatrix}$$

This is quickly seen to be both bijective and a homomorphism, so an isomorphism, explicitly identifying $G$ with the easy to understand group $G'$.

The nicest thing about this isomorphism, I think, is that it is extremly explicit. You can see the subgroups generated by $\sigma$ and $\tau$, you can easily tell which subgroups are normal and you can quickly compute the composition of given elements. This also gives you for free that the size of the extension is $p(p-1)$, although that was actually easy to compute.

Hint: Clearly $\big<\sigma \big> \big< \tau \big>=G$ (count the order of each), and $\big<\sigma \big> \cap \big< \tau \big>=e$. If one of these is normal, then it's easy to write as a semi-direct product of groups.