$
\newcommand{\fp}{{\mathfrak p}}
\newcommand{\fP}{{\mathfrak P}}
$
The ramification in intermediate fields is a slightly delicate (purely group theoretic) issue. First, assume without loss of generality that $L/K$ is Galois with Galois group $G$, otherwise pass to the Galois closure and repeat the following argument for $L$ being the intermediate field.
Suppose that you have a prime $\fp$ of $K$ with prime $\fP$ of $L$ lying above $\fp$. Let $D=D_{\fP/\fp}$ be the decomposition group, i.e. the subgroup of $G$ consisting of elements that fix the place $\fP$. Inside that, you have the inertia subgroup $I=I_{\fP/\fp}$. Recall that the index of $D$ in $G$ is the number of primes in $L$ lying above $\fp$, and $|I|$ is the ramification index of $\fP/\fp$. The residue field degree is the index of $I$ in $D$. Now, let $E$ be an intermediate field, corresponding to a subgroup $H$ of $G$, so that $H=\text{Gal}(L/E)$. Then
Exercise 1: There is a natural bijection between double cosets $H\backslash G/D$ and the primes of $E$ lying above $\fp$. (You might want to begin by checking that this is independent of the choice of $\fP$. Another prime above $\fp$ would have given a conjugate decomposition group.)
Exercise 2: If, under the above correspondence, the double coset $HgD$ corresponds to the prime $\fp'$ of $L$, then
$$
D_{\fP/\fp'} = H\cap D^g\leq H\text{ and } I_{\fP/\fp'} = H\cap I^g\leq H.
$$
This allows you to compute the ramification index and the residue field degree of $\fP/\fp'$ as above.
I hope that that answers your questions. As for explicit computation of Hilbert class fields, you might want to consult Sections 3 and 4 of Cohen's Advanced Topics in Computational Number Theory - a wonderful book that treats your question in great detail.
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.
Best Answer
You said you know about using the discriminant of a $\mathbf Z$-basis of the integers of a number field to detect ramification. You can do the same thing in a local field: if $K$ is a finite extension of $\mathbf Q_p$ then the discriminants of all $\mathbf Z_p$-bases of $\mathcal O_K$ are equal to multiplication by the square of a unit in $\mathbf Z_p$, and $K/\mathbf Q_p$ is ramified if and only if the discriminant of a $\mathbf Z_p$-basis of $\mathcal O_K$ is divisible by $p$ (equivalently, the discriminant is not a unit in $\mathbf Z_p$).