= Solution
Let $q^i=e^i-\tfrac12h^i{}_je^j$, so $p^0=\epsilon/a^2$ and $p^i=(\epsilon/a^2)q^i$. The <orthonormal tetrad> gives $\delta_{ij}e^ie^j=1$. To first order,
$$
\delta_{ij}q^iq^j=1-h_{ij}e^ie^j,\qquad h_{ij}q^iq^j=h_{ij}e^ie^j.
$$
Hence the time-index <Christoffel symbols> contract to
$$
\Gamma^0{}_{00}+\Gamma^0{}_{ij}q^iq^j=2\mathcal H+\frac12\dot h_{ij}e^ie^j.
$$
The $\mathcal H h_{ij}$ term cancels the perturbation in the coordinate direction norm. This cancellation is why the local tetrad direction must not be mistaken for the coordinate direction.
Along the <photon> <null geodesic>, $d/d\lambda=p^0d/d\eta$, and
$$
\frac{dp^0}{d\lambda}=\left(\frac\epsilon{a^2}\right)^2\left(\frac1\epsilon\frac{d\epsilon}{d\eta}-2\mathcal H\right).
$$
The time component of the <geodesic equation> now gives the <photon energy redshift from a tensor metric perturbation>:
$$
\boxed{\frac1\epsilon\frac{d\epsilon}{d\eta}=-\frac12\dot h_{ij}e^ie^j.}
$$
The derivative on the left follows the ray; the dot on $h$ is a partial <conformal time> derivative. The background redshift has already been removed by using comoving energy. The calculation only needs $\Gamma^0{}_{00}$ and $\Gamma^0{}_{ij}$; the unrelated printed $\Gamma^i{}_{j0}$ should contain $\tfrac12\dot h^i{}_j$, rather than the full derivative.
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