Let $k$ be a field of characteristic $0$ and let 
$$E: y^2 = f(x)$$
be an elliptic curve over $k$, with $\mathrm{deg}(f) = 3$. Kummer theory yields a map
$$\varphi:\mathrm{H}^1(k, E[2]) \to \mathrm{H}^1(k, E)[2].$$ In particular, to each element $c \in \mathrm{H}^1(k, E[2])$ we may associate a principal homogeneous space for $E$.



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Under the assumption that $f$ is completely split (so that $E$ has full $2$-torsion), this is done in many books. Here we have $\mathrm{H}^1(k, E[2]) \cong k^\*/k^{\*2} \times k^\*/k^{\*2}$, and given such a pair one can write down explicit equations for the associated homogeneous space.


If $f$ is irreducible, then I believe that $E[2] \cong \mathrm{R}\_{K/k}(\mu\_2)/ \mu\_2$ as group schemes, where $K/k$ is the cubic extension associated to $f$. Hence we have $\mathrm{H}^1(k, E[2]) \cong K^\*/k^\*K^{\*2}$. Given an element $c \in K^\*$, I would like to be able to write down explicit equations of the associated principal homogeneous space. I'm guessing that one would need to write down these equations in the Weil restriction of an affine space.



Explicit $2$-descent on elliptic curves
It is perhaps easier to use the isomorphism of $E[2]$ with the kernel of the 
`norm' map $R\_{K/k} \mu\_2 \to \mu\_2$, which translates into
$H^1(k, E[2]) \cong \ker(N \colon K^\times/K^{\times 2} \to k^\times/k^{\times 2})$. In this case you start with $c \in K^\times$
whose norm is a square: $N(c) = a^2$. The descent map $E(k) \to H^1(k, E[2])$
is given in these terms by $(x,y) \mapsto x-\theta$ (mod squares), where
$\theta$ is the image of $x$ in $K = k[x]/\langle f\rangle$. Now, given $c$,
we want to sort of reverse this and parametrise the preimages of $c$.
This translates into
$$ y^2 = f(x), \quad x - \theta = c (z\_0 + z\_1 \theta + z\_2 \theta^2)^2. $$
The right hand side (using $c = c\_0 + c\_1 \theta + c\_2 \theta^2$)
can be expanded as
$$ Q\_0(z\_0,z\_1,z\_2) + Q\_1(z\_0,z\_1,z\_2) \theta + Q\_2(z\_0,z\_1,z\_2) \theta^2 $$
with quadratic forms $Q\_j$ over $k$. Comparing coefficients, we obtain
$$ Q\_2(z\_0,z\_1,z\_2) = 0, \quad Q\_1(z\_0,z\_1,z\_2) = -1 $$
and $x = Q\_0(z\_0,z\_1,z\_2)$ can be eliminated; $y$ is given (up to a sign
that one has to fix) by $y^2 = N(c) N(z\_0 + z\_1 \theta + z\_2 \theta^2)^2$,
so (say) $y = a N(z\_0 + z\_1 \theta + z\_2 \theta^2)$.


Homogenizing, we obtain an intersection of two quadrics in $\mathbb P^3$:
$$ C \colon \quad Q\_2(z\_0,z\_1,z\_2) = 0, \quad Q\_1(z\_0,z\_1,z\_2) + z\_3^2 = 0 ,$$
which is a model of the principal homogeneous space you are looking for.
(You also get the 2-covering map to $E$ via $x$ and $y$ as above.)


Note that when $k$ is a number field and $C$ has points over all completions
of $k$, then the conic given by the first equation has a $k$-point and
can be parametrised by binary quadratic forms. Plugging this into the
second equation gives you the $y^2 = \text{quartic}$ model.


