= Solution
Treat the <electron> and <electron antineutrino> as effectively <massless particles> in the <centre-of-momentum frame>, retain only the <charged current>, and use the <Fermi interaction>: $m_e^2\ll s\ll M_W^2$. Take $\theta^*$ between incoming and outgoing <electron antineutrino> directions. The <V-A interaction> selects a right-helicity <electron antineutrino> and a left-helicity <electron>. Their opposite incoming directions give total <angular momentum> projection one. The outgoing pair has the same projection along its own axis. The <Wigner D-matrix> element is $d^1_{11}(\theta^*)=(1+\cos\theta^*)/2$, so the squared <helicity amplitude> is proportional to $(1+\cos\theta^*)^2$. The inactive initial <electron> <helicity> is included in the unpolarized normalization supplied by the paper.
The boost from the <centre-of-momentum frame> to the <laboratory frame> has $\beta=E/(E+m_e)$, while the incoming <electron antineutrino> has $E_*=(s-m_e^2)/(2\sqrt s)=m_eE/\sqrt s$ and $s=m_e^2+2m_eE$. Thus
$$
\frac{E'}E=\frac{E+m_e+E\cos\theta^*}{2E+m_e},\qquad y=\frac{E(1-\cos\theta^*)}{2E+m_e}\simeq\frac{1-\cos\theta^*}{2}.
$$
The requested identity therefore uses $E\gg m_e$; it is not exact at finite target <mass>. With $c_*=\cos\theta^*$, the absolute <Jacobian determinant> is $|dy/dc_*|=1/2$, giving
$$
\frac{d\sigma}{dc_*}=\frac{G_F^2s}{8\pi}(1+c_*)^2,\qquad
\frac{d\sigma}{d\Omega_*}=\frac{G_F^2s}{16\pi^2}(1+c_*)^2,\qquad
\sigma=\frac{G_F^2s}{3\pi}.
$$
The suppression at backward <scattering> follows from <angular momentum> conservation: rotating projection $+1$ through $\pi$ cannot leave projection $+1$ along the original axis. Replacing $1+c_*$ by $2(1-y)$ gives the supplied $(1-y)^2$ dependence.
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