= Solution
Use metric compatibility and the symmetric <Levi-Civita connection>. Expanding $k_a=g_{ac}k^c$ gives
$$
\nabla_bk_a+\nabla_ak_b
=g_{ac}\partial_bk^c+g_{bc}\partial_ak^c+
k^c(\partial_bg_{ac}+\partial_ag_{bc}-2\Gamma^d_{ab}g_{dc}).
$$
The Christoffel formula makes the bracket $\partial_cg_{ab}$. Thus the <Killing equation> is equivalent to the displayed coordinate formula in the question, also the vanishing <Lie derivative> of the metric.
Along an affinely parametrized <geodesic> with tangent $V$,
$$
V^b\nabla_b(V^ak_a)=k_aV^b\nabla_bV^a+V^aV^b\nabla_bk_a=0.
$$
The first term vanishes by the geodesic equation and the second by the Killing equation, because $V^aV^b$ is symmetric. This proves the <geodesic conserved quantity from a Killing vector>.
For $ds^2=-du^2+u^2dv^2$, the three independent coordinate Killing conditions are
$$
\partial_uk^u=0,\qquad uk^u+u^2\partial_vk^v=0,\qquad-\partial_vk^u+u^2\partial_uk^v=0.
$$
The field $(0,1)$ satisfies all three immediately. For $e^{-v}(1,u^{-1})$, the second equation reads $ue^{-v}-ue^{-v}=0$, and the third reads $e^{-v}-e^{-v}=0$. For $e^v(-1,u^{-1})$, they read $-ue^v+ue^v=0$ and $e^v-e^v=0$. The first equation holds for both, verifying all three <Killing vector fields>.
Their conserved contractions with $(\dot u,\dot v)$ may be denoted
$$
\gamma=u^2\dot v,\qquad
\alpha=e^{-v}(-\dot u+u\dot v),\qquad
\beta=e^v(\dot u+u\dot v).
$$
Adding the last two expressions after multiplying by $e^v,e^{-v}$ gives
$$
\boxed{\alpha e^v+\beta e^{-v}=2\gamma/u}.
$$
To identify these loci, set $T=u\cosh v$, $X=u\sinh v$. The metric becomes $-dT^2+dX^2$, and the equation becomes $(\alpha+\beta)T+(\alpha-\beta)X=2\gamma$, a straight line in <Minkowski spacetime>. Nontrivial such lines, parametrized affinely, are exactly the geodesics in this coordinate patch. The all-zero choice of constants is a vacuous equation rather than an individual geodesic.
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