= Solution
The <orthogonal projection> onto the full spin-$\eta$ subspace is
$$
\Pi_\eta=|\eta\rangle\langle\eta|\otimes I_{\mathrm{osc}}
=\sum_{\{n'\}}|\eta,\{n'\}\rangle\langle\eta,\{n'\}|.
$$
Here the identity acts on all oscillator states. The finite-time <scattering operator> $S(t)$ in the question is the <interaction picture> <time-evolution operator>, so $|\psi_\sigma^I(t)\rangle=S(t)|\sigma,\{n\}\rangle$. Projecting its two orthogonal spin components yields
$$
\boxed{c_{\eta\sigma}(t)|\phi_\eta(t)\rangle
=\Pi_\eta S(t)|\sigma,\{n\}\rangle.}
$$
Expand the right-hand side in the <orthonormal basis> $|\eta,\{n'\}\rangle$. Taking its squared <norm>, and using the normalization of $|\phi_\eta\rangle$, gives
$$
\boxed{|c_{\eta\sigma}(t)|^2
=\sum_{\{n'\}}|\langle\eta,\{n'\}|S(t)|\sigma,\{n\}\rangle|^2
=\sum_{\{n'\}}P_{\sigma,\{n\}\to\eta,\{n'\}}.}
$$
Thus one sums <probabilities> over unobserved final oscillator states, not their <probability amplitudes>. <Unitarity> of $S(t)$ makes the sum over both $\eta$ equal to one.
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