I was hoping for a condition in terms of arithmetic of the field, e.g. existence of a finite field extension of a certain type.

Thanks, Jason, that's great! I see that the projection of $X$ to the the first factor $\mathbf{P}^2$ identifies $X$ with a blow up of $\mathbf{P}^2$ in $d^2$ points obtained from a pair of intersecting degree $d$ plane curves, in particular $X$ is rational and so $p_g(X) = q(X) = 0$. Furthermore, I see how for large $d$, $\chi(L) < 0$, hence $L$ has nonvanishing $h^1$.

I wonder if a single uniruled parametrization may ever suffice in this question. Namely, do there exist uniruled parametrizations with images of all components of all fibers being singular rational curves? It's easy to arrange such parametrization with general curve being singular, but these typically break up to have smooth components.

For singular curves, the nontrivial part of $K_0(X)$ should come from the Picard group $Pic(X)$ which can be computed in terms of the local structure of singular points. One reference about Picard group of singular curves is link.springer.com/content/pdf/10.1007/BF01440949.pdf.

This version of the Luroth problem for the fields of invariants is sometimes called the Noether problem: encyclopediaofmath.org/wiki/….

Higher degree unirational parametrizations of cubics with a rational point are explained in arxiv.org/pdf/math/0005146.pdf.

Geometrically, the question translates into degree of unirationality of this surface. Namely, if there is a degree two dominant rational map from $\mathbf{P}^2$ to this cubic, then the corresponding field extension would automatically be Galois. Typically, degree two parametrization of cubic hypersurfaces comes from a line defined over a ground field, as in math.uni-bonn.de/~huybrech/Notes.pdf, Corollary 1.18. However in this case the only line I see (at infinity) is passing through singular points, and the construction doesn't work.

(Q1): If $X$ is a smooth projective curve, and $G$ a finite group acting on it, then $X/G$ is smooth projective curve. Using Riemann-Hurwitz formula one shows that Q1 can only hold when $X$ has genus $0$ (always) or genus $1$.

An excellent modern survey by Beauville: arxiv.org/pdf/1507.02476.pdf; see section 3 for the intermediate Jacobian.

If the curve has a nontrivial finite group action, the induced action on the Jacobian can split off an elliptic curve. This will be the case if a group element has an eigenvalue 1 of multiplicity one on the tangent space to the Jacobian (which is the dual to the space of 1-forms on the curve).

The title of the question has general type in it, but not the question itself! Without general type assumption, e.g. for del Pezzo surfaces, the statement will be definitely false because the canonical class of the fibers will be negative, so the pushforward will be zero, but del Pezzo surfaces have moduli.

At least on the level of points of the moduli space, to specify a rational normal curve of degree $r$, you can first parametrize it as a map from $\mathbf{P}^1$ to $\mathbf{P}^r$ of degree $r$ (Veronese embedding). This amounts to choosing the basis of degree $r$ polynomials in $X$, $Y$. Two maps give the same curve if they differ by linear action on $X$, $Y$.

You can represent the open subscheme of the Hilbert scheme as a GIT quotient. For instance for twisted cubics, one needs to choose 4 homogeneous polynomials of degree $3$ in $X$, $Y$, up to common scaling and then quotient out the $\mathbf{PGL}_2$-action on $X$, $Y$. This gives a $12$-dimensional component of the Hilbert scheme.

In characteristic zero, classes of fields are linearly independent by the Larsen-Lunts theorem, as in the answer here: mathoverflow.net/questions/354718/…

(1) is a bit strange: RHS is a scheme, so why do you ask for isomorphism of stacks? The natural question is whether $K_0(L)^G$ is isomorphic to $K_0(K)$.