There is a typo in the statement of Lemma 67 in your source. The $n$th roots of unity are in $\Bbb{F}_p$ only if $n\mid p-1$ or, iff $p\equiv1\pmod n$ (not $n\equiv1\pmod p$ as is written there). Therefore that Lemma does not apply.
In fact, the polynomial $x^{15}-2$ is NOT irreducible in $\Bbb{F}_7[x]$. This follows trivially from the fact that $3^5=243\equiv-2\pmod 7$. Therefore
$$
x^{15}-2=(x^3)^5+3^5=(x^3+3)(x^{12}-3x^9+3^2x^6-3^3x^3+3^4).
$$
We immediately see that $x^3+3$ has no zeros in $\Bbb{F}_7$ (the cubes in that field are $0,\pm1$), so it is irreducible. Therefore the polynomial has a zero $\alpha$ in $\Bbb{F}_{7^3}$.
To get the splitting field of $x^{15}-2$ we need, as you observed, the primitive 15th roots of unity. We easily see that
$$
7^4=2401\equiv1\pmod{15}.
$$
The multiplicative group of the field $\Bbb{F}_{7^4}$ is cyclic of order $7^4-1$, and thus it contains a primitive 15th root of unity $\zeta$.
A consequence of all this is that the splitting field of this polynomial is
$$
\Bbb{F}_7[\alpha,\zeta]=\Bbb{F}_{7^{12}}.
$$
Let $ H_i $ denote the isomorphic copy of $ S_{n-1} $ in $ S_n $ composed of elements which fix $ i $. For $ L/K $ a Galois extension with Galois group $ S_n $, consider the fixed field $ F_1 $ of $ H_1 $. Clearly $ [F_1 : K] = n $, and $ F_1/K $ is separable since it is a subextension of the separable extension $ L/K $. Pick a primitive element $ \alpha_1 $ for this extension. Then, $ \alpha_1 $ has exactly $ n $ $ K $-conjugates (counting itself), say $ \alpha_1, \alpha_2, \ldots, \alpha_n $, since its stabilizer in the Galois group is exactly $ H_1 $, which is a subgroup of index $ n $ in the Galois group. The subgroup of $ S_n $ fixing $ K(\alpha_1, \alpha_2, \ldots, \alpha_n)/K $ is exactly
$$ \bigcap_{1 \leq i \leq n} H_i = \{ \textrm{id} \} $$
By Galois correspondence, it follows that $ K(\alpha_1, \alpha_2, \ldots, \alpha_n) = L $. Therefore, $ L $ is the splitting field of the minimal polynomial of $ \alpha_1 $ over $ K $, which is a polynomial of degree $ n $.
Best Answer
Your first question should be: is this polynomial irreducible in $\Bbb F_3[x]$?
It's clear it has no root in $\Bbb F_3$, but this is not enough, we must check for possible quadratic factors. So, suppose (by way of contradiction) we had:
$x^4 + x - 1 = (x^2 + ax + b)(x^2 + cx + d)$ with $a,b,c,d \in \Bbb F_3$.
Then $a + c = 0$ (since our polynomial has no cubic term), and $bd = -1$. The latter equation implies either $b = 1, d = 2$, or $b = 2, d = 1$.
We also have $b + d - a^2 = 0$, and $a(d - b) = 1$. If $d = 2, b = 1$, then $a(d - b) = 1$ implies that $a = 1$ in which case $b + d - a^2 = 1 + 2 - 1 = 2 \neq 0$.
On the other hand, if $b = 2, d= 1$, then $a(d - b) = 1$ implies that $a = 2$, and thus $b + d - a^2 = 2 + 1 - 1 = 2\neq 0$. Either way, we see we can have no quadratic factors, and our polynomial is indeed irreducible.
Hence $\Bbb F_3(u)$ is a field of dimension $4$ over $\Bbb F_3$ (as a vector space), for any root $u$ of $x^4 + x - 1$. Now $\Bbb F_3$ is a perfect field, so this is a Galois extension of $\Bbb F_3$. Thus we know that: $|\text{Gal}(\Bbb F_3(u)/\Bbb F_3)| = \dim_{\Bbb F_3}(\Bbb F_3(u)) = 4$.
Furthermore, it is not hard to see that:
$\sigma: u \mapsto u^3 \in \text{Gal}(\Bbb F_3(u)/\Bbb F_3)$.
Since $\sigma^2 \neq \text{id}_{\Bbb F_3(u)}$, we see this automorphism is of order $4$ and the Galois group is cyclic.