= Solution
Let $\Delta_M$ be the free real-scalar <Feynman propagator> and $D_m(x-y)=\langle0|T\varphi(x)\varphi^\dagger(y)|0\rangle_0$ the free complex-scalar one. Expanding the interaction exponential once and applying <Wick's theorem> gives the connected decay <Green's function>
$$
\boxed{\langle T\phi(x)\varphi(y)\varphi^\dagger(z)\rangle_c
=-ig\int d^4w\,\Delta_M(x-w)D_m(y-w)D_m(w-z)+O(g^3).}
$$
There is one allowed contraction of each external <field> with the appropriate <vertex> field, so no factorial remains. Fourier transformation gives the momentum-conserving delta factor times $(-ig)\widetilde\Delta_M(p)\widetilde D_m(q_1)\widetilde D_m(q_2)$, and <amputation of external propagators> gives the decay vertex $-ig$. An on-shell physical decay additionally requires $M>2m$.
If one literally asks for the full three-point <correlation function> without a connected restriction or a tadpole condition, it also contains
$$
-ig\int d^4w\,\Delta_M(x-w)D_m(0)D_m(y-z),
$$
the product of the induced heavy-field one-point function and the free light-field two-point function. This disconnected <tadpole diagram> does not describe a three-particle decay and drops out of its connected amputated amplitude.
For two equally charged daughters the corresponding <Green's function> vanishes to every perturbative order:
$$
\boxed{\langle T\phi(x)\varphi^\dagger(y)\varphi^\dagger(z)\rangle=0.}
$$
Every interaction contains equal numbers of $\varphi$ and $\varphi^\dagger$, so the two excess conjugate fields can never be paired. Equivalently the exact global <U(1)> symmetry $\varphi\mapsto e^{i\alpha}\varphi$, $\varphi^\dagger\mapsto e^{-i\alpha}\varphi^\dagger$ leaves the <vacuum> and action invariant, whereas this insertion has charge $-2$. Charge conservation forbids the decay of a neutral <particle> into two particles of the same nonzero charge.
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