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For a relativistic species at zero chemical potential, $p=ϵ/3$ and $s=(ϵ+p)/T_i$, so its entropy density is proportional to $g_i{T}_i^3$. The Bose-Einstein distribution gives the bosonic normalization, while the fermionic thermal integral is $7/8$ of it. Species still in thermal equilibrium have $T_i=T$; freely streaming decoupled species can have another temperature and therefore contribute an extra factor $(T_i/T{)}^3$ when the total is expressed in units of $T^3$. This gives the two terms $N_{\ast }$ and $N_{SD}$ and their stated effective-degree-of-freedom weights.

After neutrino decoupling, the comoving entropy of the photon-electron-positron plasma is conserved separately from the neutrinos. Before electron-positron annihilation its effective entropy count is $2+(7/8)4=11/2$; afterwards it is $2$. Consequentlywhereas $a{T}_{\nu }$ stays equal to its pre-annihilation value. HenceThis is the instantaneous-decoupling, negligible residual-electron approximation used in the question. Heating of the still-coupled plasma, not cooling peculiar to neutrinos, creates the temperature difference.