= Solution
The authoritative PDF has $m_Z^2/v$, not the dimensionally incorrect $m_Z^2/v^2$ in the TeX conversion. The two identical <Z bosons> give the vertex factor two, so the <scattering amplitude> for the decay is
$$
\mathcal M=\frac{2m_Z^2}{v}\,\epsilon_1^*\cdot\epsilon_2^*,
$$
up to an irrelevant overall phase. Apply the <massive vector polarization sum>, identifying the hint's $M_Z$ with $m_Z$:
$$
\sum_{s_1,s_2}|\epsilon_1\cdot\epsilon_2|^2=\left(-g_{\mu\nu}+\frac{k_{1\mu}k_{1\nu}}{m_Z^2}\right)\left(-g^{\mu\nu}+\frac{k_2^\mu k_2^\nu}{m_Z^2}\right)=2+\frac{(k_1\cdot k_2)^2}{m_Z^4}.
$$
Since $k_1\cdot k_2=(m_H^2-2m_Z^2)/2$, putting $x=m_Z^2/m_H^2$ gives
$$
\sum_{\mathrm{spins}}|\mathcal M|^2=\frac{m_H^4}{v^2}(1-4x+12x^2).
$$
No initial-spin average is needed for a <Higgs boson>. In its rest frame the two-body momentum is $|\boldsymbol k|=(m_H/2)\sqrt{1-4x}$. Integrating the energy and momentum delta functions in the <Lorentz-invariant phase-space measure> gives $d\Phi_2=\sqrt{1-4x}\,d\Omega/(32\pi^2)$, hence $\int d\Phi_2=\sqrt{1-4x}/(8\pi)$.
The printed generic width formula treats the daughters as labelled. Here the <identical final-state symmetry factor> is $1/2!$, without which the same physical configuration is counted twice. Thus the <Higgs decay to two Z bosons> has
$$
\boxed{\Gamma(H\to ZZ)=\frac{m_H^3}{32\pi v^2}\sqrt{1-4x}(1-4x+12x^2),\qquad m_H>2m_Z.}
$$
Equivalently, $1/v^2=\sqrt2G_F$ gives the prefactor $G_Fm_H^3/(16\sqrt2\pi)$. The threshold limit is zero and the large-mass limit is $m_H^3/(32\pi v^2)$. The additional factor for identical daughters is a required specialization of the PDF's generic phase-space formula, not an extra factor in the vertex.
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