= Solution
There are two bifurcations in $0\leq k\leq1/2$, both at $k=0$.
Near $(0,0,0)$, append $\dot k=0$. The plane $y=0$ is invariant and tangent to the extended center subspace, so it is an <extended centre manifold for a parameter>[extended center manifold]. The reduced equation is exactly
$$
\dot x=x(-k-3x+x^2),
\qquad
\dot k=0.
$$
Its leading terms $-kx-3x^2$ have two branches, $x=0$ and $x=-k/3+O(k^2)$, which cross and exchange their center-direction stability. Hence this is a <transcritical bifurcation>.
For the bifurcation at $(1,2,0)$, set
$$
X=x-1,\qquad v=y-2-X.
$$
The extended system becomes
$$
\dot X=(1+X)(v-k+X^2),
\qquad
\dot v=v+v^2+(1+X)(k-X^2),
\qquad
\dot k=0.
$$
Solving the <centre-manifold invariance equation> for $v=h(X,k)$ gives
$$
h(X,k)=-k+X^2-5Xk+9k^2+O\bigl((|X|+|k|)^3\bigr).
$$
Substitution into the $X$ equation yields
$$
\dot X=2(X^2-k)+O\bigl((|X|+|k|)^3\bigr).
$$
For $k>0$ this has two nearby equilibria $X=\pm\sqrt k+O(k)$, while for $k<0$ it has none. It is therefore a <saddle-node bifurcation>. These reductions are collected in <bifurcations of the 2022 Cambridge quadratic-cubic system>.
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