This is an exercise I’ve never done, but it should be a lot of fun. What is the general Eisenstein polynomial in this case? it’ll be
$$
X^3 + 2aX^2+2bX+2(1+2c)\,,
$$
where $a$, $b$, and $c$ can be any $2$-adic integers. Notice that the constant term has to be indivisible by any higher power of $2$, so of form $2$ times a unit, and the units of $\Bbb Z_2$ are exactly the things of form $1+2c$. So your parameter space is $\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_2$, pleasingly compact, and a general result of Krasner says that if you jiggle the coefficients a little, the extension doesn’t change. You might be able to use all this to construct your (finitely many) fields.
Not much of an answer, I know, but it was too long for a comment. It’s a nice question, though, and I think I’m going to worry it over a little.
EDIT — Expansion:
I told no lies above, but that’s not the way to look at this problem. As I reached the solution, I realized that there are really two questions here. Consider the simplest case, which you mentioned, the Eisenstein polynomial $X^3-2$. If you think of it abstractly, there’s only the one extension of $\Bbb Q_2$ here, but if you think of the subfields of some algebraically closed containing field, there are three fields, generated by $\lambda$, $\omega\lambda$, and $\omega^2\lambda$, where $\lambda$ is a chosen cube root of $2$ and $\omega$ is a primitive cube root of unity.
As usual, if you take the displayed cubic above and make a substitution $X'=X-2a/3$, you’ll get a new Eisenstein polynomial, but without a quadratic term. Now, if you calculate the discriminant of $X^3+2bX+2(1+2c)$, you’ll get $\Delta=-32b^3-27(1+4c+4c^2)$; and since $c$ and $c^2$ have same parity, we get $\Delta\equiv-3\pmod8$, definitely not a square, indeed $\sqrt\Delta\in\Bbb Q(\omega)$, hardly a surprise, I suppose. And the splitting field of our polynomial will be a cubic extension of $k=\Bbb Q_2(\omega)$, all of which we know. We need only calculate the group $k^*/(k^*)^3$, and its cyclic subgroups (of order $3$) will tell us the cubic extensions of $k$. That’s Kummer Theory, as I’m sure you know.
Let’s call $\Bbb Z_2[\omega]=\mathfrak o$, that’s the ring of integers of $k$.
To know $k^*/(k^*)^3$ we have to look at the groups $1+2\mathfrak o\subset \mathfrak o^*\subset k^*$. Now the principal units $1+2\mathfrak o$ are uniquely $3$-divisible, so no contribution to $k^*/(k^*)^3$; the next layer, $\mathfrak o^*/(1+2\mathfrak o)$ is cyclic of order $3$, generated by $\omega$, and $k^*/\mathfrak o^*$ is infinite cyclic, that’s the value group. So $k^*/(k^*)^3$ is of dimension two as an $\Bbb F_3$-vector space, and has only four one-dimensional subspaces. One is spanned by $\omega$, and its cube roots generate an unramified extension, so is not of interest to us. The other three are spanned by $2$, $2\omega$, and $2\omega^2$. ( ! )
And that’s it. Contrary to my expectation and perhaps yours, the only cubic ramified extensions of $\Bbb Q_2$ within an algebraic closure are the three I mentioned in the first paragraph of this Edit.
$q$ is prime, $L/K$ is cyclic of degree $q^r$, $L'/K$ is unramified of degree $q^{r-s}>1$ and $L/L'$ is totally ramified of degree $q^s>1$, then any automorphism $\sigma\in Gal(L/K)$ such that $\sigma|_{L'}$ is the Frobenius will be a generator of $Gal(L/K)$.
Since $F/K$ is unramified of degree $q^r$ then
$[LF:L]=[F:L'] = q^s$ and $[LF:K] = q^r q^s$
$$Gal(L/L')\times Gal(F/L') \cong Gal(LF/L') = \langle a\rangle \times \langle b\rangle = C_{q^s}\times C_{q^s}$$
where $b$ is the Frobenius of $LF/L$ and $a$ is a generator of $Gal(LF/F)$.
Let $\phi$ be an extension of the Frobenius of $F/K$ to $LF$, then $$\phi^{q^{r-s}}|_F=b|_F$$ so $\phi^{q^{r-s}} = a^n b$, the order of $\phi$ is $q^r$ so that $\langle a\rangle \cap \langle \phi \rangle = \{1\}$ and $$
Gal(LF/K) = \langle a\rangle \times \langle \phi \rangle = C_{q^s} \times C_{q^ r}$$
Best Answer
The answer is no, even for Galois extensions. The standard example is as follows. Suppose that $L/K$ is Galois with $\mathrm{Gal}(L/K) = \mathbf{Z}/4 \mathbf{Z}$ is cyclic of order $4$ but the inertia group is cyclic of order $2$, so $f = e = 2$. By Galois theory, there is only one (proper) intermediate subfield $L/E/K$, and it will be unramified over $K$, so $L/E$ will be ramified. You can construct examples of such extensions quite explicitly. Suppose that $p \equiv 1 \bmod 4$ so that $\mathbf{Q}_p$ contains a $4$th root of unity $i$. Now let $u \in \mathbf{Z}_p$ be any element which is a quadratic non-residue. Then take $K = \mathbf{Q}_p$ and $L = \mathbf{Q}_p((up^2)^{1/4})$.
For Galois extensions, with Galois group $G$, inertia group $I$, and cyclic quotient $G/I = \langle \phi \rangle$, the general condition is that there exists a splitting $\langle \phi \rangle \rightarrow G$, that is, there is a lift of $\phi$ to $G$ which has the same order. Then you can take $E$ to be the fixed field of that lift $\phi \subset G$. You can see the example above is non-split.