= Solution
Use one <Fourier mode>, with $\mu=\widehat{\mathbf k}\cdot\widehat{\mathbf n}$ and unperturbed ray $\mathbf x(\tau)=\widehat{\mathbf n}(\tau_0-\tau)$. Put $E(\tau)=e^{ik\mu(\tau_0-\tau)}$, and let $\tau_*$ denote last scattering. In the supplied branch $h'=h_s'$, the scalar decomposition gives
$$
h_{ij}'n_in_j=h'/3+h_s'(\mu^2-1/3)=\mu^2h_s'.
$$
Since $E''=-k^2\mu^2E$, two <integrations by parts> give the exact endpoint decomposition of the metric contribution:
$$
-\frac{\mu^2}{2}\int_{\tau_*}^{\tau_0}h_s'E\,d\tau
=\frac{[h_s'E'-h_s''E]_{\tau_*}^{\tau_0}}{2k^2}
+\frac1{2k^2}\int_{\tau_*}^{\tau_0}h_s'''E\,d\tau.
$$
For the matter-era growing mode, $h_s$ is quadratic in $\tau$, so the last integral vanishes. With $E'=-ik\mu E$, the emission endpoint is
$$
E_*\left(\frac{h_s''}{2k^2}+\frac{i\mu h_s'}{2k}\right)_*,
$$
and the observer endpoint is $-h_s''(\tau_0)/(2k^2)-i\mu h_s'(\tau_0)/(2k)$. The observer terms are only a <angular monopole> and <angular dipole>; they are removed when studying the anisotropy multipoles $\ell\geq2$.
For an adiabatic growing mode outside the horizon at last scattering, $\delta_\gamma=4\delta_m/3$, $\delta_m=-h/2$ after fixing the constant trace convention, and $h\propto\tau^2$. Hence the intrinsic emission term is smaller than $h_s''/(2k^2)$ by order $(k\tau_*)^2$. The derivative emission endpoint is smaller by order $k\tau_*$. <Stress-energy conservation> also makes the peculiar emission velocity subleading; its leading synchronous contribution cancels in $\delta_\gamma'+2h'/3$, with gradients supplying the remaining small term. Keeping the Fourier phase, which need not be small over the observer's distance, gives
$$
\boxed{\Theta(\widehat{\mathbf n})\simeq\frac{h_s''(\tau_*)}{2k^2}e^{i\mathbf k\cdot\widehat{\mathbf n}\chi_*},\qquad \chi_*=\tau_0-\tau_*.}
$$
Here $\Theta=\delta T/T$, and the sum over modes is understood. The derivative is the \b[second derivative of $h_s$], as printed in the original PDF. This approximation assumes matter-era growing adiabatic modes and neglects late evolving potentials; large wavelength alone does not remove the last integral in a general cosmology. The equality $h'=h_s'$ is used as the supplied synchronous condition, rather than as a general identity for every fluid solely from vanishing anisotropic stress.
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