Yes, this is (provably!) dominant.
As Jackson Morrow pointed out, quadratic points on $C$ are the same as rational points on the symmetric square of $C$. The obvious points are the rational points on a $\mathbb P^1$. Contracting this $\mathbb P^1$, we get a surface in an abelian variety, hence by Falting's theorem (as Jackson Morrow also pointed out), contained in finitely many abelian subvarieties plus finitely many points.
For each abelian variety, the number of points of height at most $X$ will be $O ((\log X)^{r/2})$ where $r$ is the rank. On the other hand, $\mathbb P^1$ has $\approx X^2$ points of height $<X$.
So we can see that the obvious points are dense, as soon as we check that the heights in $\operatorname{Sym}^2 C$ and $C$ are comparable. But this is straightforward - we can map $\operatorname{Sym}^2 C \to \operatorname{Sym}^2 \mathbb P^1 \to \mathbb P^2$ by $( (x_1:y_1),(x_2:y_2)) \to ( x_1x_2: x_1 y_2+ x_2 y_1 : x_2 y_2)$$( (x_1:y_1),(x_2:y_2)) \to ( x_1x_2: x_1 y_2+ x_2 y_1 : y_1 y_2)$ which makes it clear that if $(x_1:y_1)$ and $(x_2:y_2)$ are conjugate and have the same Weil height, their heights are at least the square root of the height of $( x_1x_2: x_1 y_2+ x_2 y_1 : x_2 y_2)$$( x_1x_2: x_1 y_2+ x_2 y_1 : y_1 y_2)$.