= Solution
For an affine point of $C$ with $x\ne0$, the first coordinate of its image is $(y/x)^2$. Thus its image has trivial <square class> whenever that coordinate is nonzero. The identity and $T=(0,0)$ map to the identity. If a point maps to $T_1=(0,0)$, it has $y=0$ and $x^2+ax+b=0$, forcing $b_1=a^2-4b$ to be a rational square; in that case $q(T_1)=1$ too. Hence $\phi(C(\mathbb Q))\subseteq\ker q$.
Conversely, let $(u,v)\in D(\mathbb Q)$ have $q(u,v)=1$ and $u\ne0$. Write $u=t^2$ with $t\in\mathbb Q^*$. The equation of $D$ gives
$$
(v/t)^2=(u-a)^2-4b.
$$
Set $x=(u-a+v/t)/2$ and $y=tx$. Then $x$ solves $x^2+(a-u)x+b=0$ and is nonzero because $b\ne0$. Hence $y^2=ux^2=x(x^2+ax+b)$, so $(x,y)\in C(\mathbb Q)$. Also $x-b/x=v/t$, giving $\phi(x,y)=(u,v)$.
The identity is an image of the identity. Finally, $T_1\in\ker q$ exactly when $b_1$ is a rational square; then either nonzero root $x=(-a\pm\sqrt{b_1})/2$ gives $(x,0)\in C(\mathbb Q)$ mapping to $T_1$. This covers all exceptional points and proves
$$
\boxed{\ker q=\phi(C(\mathbb Q)).}
$$
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