The Galois group of $F/\mathbb Q$ is cyclic of order $6$ as you described above, and if $\sigma$ is a generator for this group, then since the unique subgroup of order $2$ is $\{1, \sigma^2, \sigma^4 \}$, we can assume without loss of generality that $\sigma^2 = u$. If you compute $\sigma$, you can compute $\sigma^2$ and explicit $E$ as the fixed field of $\sigma^2$.
Now the extension $F/\mathbb Q$ is simple, so $\sigma$ is characterized entirely by where it maps $f$ (let me write $f = \zeta_7$ for clarity). It suffices to find a primitive root of $(\mathbb Z/7\mathbb Z)^{\times}$ and we got it. After some testing you realize that $3^6 = 1$ but $3^2, 3^3 \neq 1 \pmod 7$, so $3$ is a primitive root. This means that $\sigma(\zeta_7) = \zeta_7^3$ is a possibility for $\sigma$ as a generator of the Galois group.
Now $\sigma^2(\zeta_7) = \zeta_7^9 = \zeta_7^2$ (as promised by $\sigma^2 = u$ and $u(f) = f^2$), so we're trying to find the fixed field corresponding to the group $\{ 1,\sigma^2,\sigma^4\}$.
Because of the simple relation $\zeta_7^7 = 1$, it is not hard to solve the equation $\sigma^2(x)-x = 0$ for $x \in F$ ; write
$$
x = a_0 + a_1 \zeta_7 + \cdots + a_5 \zeta_7^5
$$
and apply $\sigma^2$ to it ; the equation $\sigma^2(x) = x$ will give you linear conditions on the coefficients. I'm afraid finding generators involves linear algebra. If you have the patience of writing down the equation by hand, you'll see the computations do not require any smart-ideas and you get
$$
E = \mathbb Q(\zeta_7 + \zeta_7^2 + \zeta_7^4).
$$
The minimal polynomial can be computed, it is
$$
(X - (\zeta_7 + \zeta_7^2 + \zeta_7^4))(X - (\zeta_7^3 + \zeta_7^5 + \zeta_7^6)) = X^2 + X + 2.
$$
The roots of this polynomial are $\frac{-1 \pm \sqrt{-7}}2$, so you can write
$$
E = \mathbb Q(\sqrt{-7}).
$$
Note : The magical reason why I get only one canonical element (i.e. $\zeta_7 + \zeta_7^2 + \zeta_7^4$) as a generator for $E/\mathbb Q$ is because $[E:\mathbb Q] = 2$, so I know the extension can only be generated by one element and that a basis for $E/\mathbb Q$ will be given by $\{1,x\}$ for some $x \in E$. This means when I tried to compute the fixed field of $\sigma^2$, I know the term for $1$ would be irrelevant and the rest would only give me one term.
Interesting remark : You know that $\mathbb Z / 6 \mathbb Z$ has only one subgroup of index $2$ and that the extension $F/\mathbb Q$ is Galois, so by the Galois correspondence $F$ has only one subfield of order $2$ over $\mathbb Q$. The element $\zeta_7 + \sigma^2(\zeta_7) + \sigma^4(\zeta_7)$ is invariant by $\sigma^2$ and is not in $\mathbb Q$, so you could've skipped the linear algebra computations and know already that $E = \mathbb Q(\zeta_7 + \zeta_7^2 + \zeta_7^4)$. We don't escape the computation of the minimal polynomial to simplify this though. (These ideas of averaging are often used in invariant theory ; I averaged the orbit of $\zeta_7$ over the subgroup $\{1,\sigma^2,\sigma^4\}$ to find the invariant substructure I wanted.)
Hope that helps,
Hint:
Let $\mu_n$ denote the set of $n$-th roots of unity. If $n,m$ are coprime, then $\mathrm{Gal}(\mathbb{Q}(\mu_{nm})/\mathbb{Q})\cong\mathrm{Gal}(\mathbb{Q}(\mu_n)/\mathbb{Q})\times\mathrm{Gal}(\mathbb{Q}(\mu_m)/\mathbb{Q})$ via restrictions. Using this, you can find a Galois extension with Galois group $\mathbb{Z}/6\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z}$ and then you can conclude via the fundamental theorem of Galois theory. (It may be worthwhile to note that if you know the sturcture theorem for finite abelian groups and a special case of Dirichlet's theorem on arithmetic progressions, then you can prove that any abelian group can be realized as Galois group over $\mathbb{Q}$ using this method).
Best Answer
Using the fundamental theorem of Galois theory here sounds like a good idea to me. Let $K = \mathbb{Q}(b, \zeta)$ in your notation. Since $P$ is a normal subgroup of $G$, $E := K^{P}$ is a Galois extension of $\mathbb{Q}$ with $\mathrm{Gal}(E/\mathbb{Q}) \cong G/P$. Moreover, $E$ is the unique extension of $\mathbb{Q}$ contained in $K$ satisfying $[E:\mathbb{Q}] = p-1$. Indeed, for any subfield $E' \subset K$ with $[E':\mathbb{Q}] = p-1$, then writing $E' = K^{H}$ for some subgroup $H$ of $G$, we see that $[G:H] = [E':\mathbb{Q}] = p-1$, which implies that $|H| = p$, i.e. $H$ is another Sylow $p$-subgroup of $G$. This forces $P = H$, since $P$ is the unique Sylow $p$-subgroup of $G$, and so $E' = E$.
Thus, it suffices to exhibit a degree $p-1$ extension of $\mathbb{Q}$ contained in $K$ which has cyclic Galois group over $\mathbb{Q}$, since this must be the equal to $E$ by the reasoning above. I leave to you to check that $\mathbb{Q}(\zeta)/\mathbb{Q}$ is one such extension.