= Solution
Use a local <orthonormal tetrad> and units in which the <speed of light> is one. A particle has <four-momentum> $p^{\hat a}=E(1,\mathbf e)$ and moves with <velocity> $\mathbf e$. Thus its <energy density> contributes $E$, its <momentum density> contributes $Ee^i$, and its <momentum flux> contributes $Ee^ie^j$. Integrating the <phase-space distribution function> gives the <kinetic stress-energy tensor>. The measure $d^3p/E$ is the future mass-shell <Lorentz-invariant phase-space measure>, so the same expression transforms as a spacetime <stress-energy tensor>; fixed state-counting or polarization factors are understood to be included in $f$.
Define $\langle F\rangle_\Omega=(4\pi)^{-1}\int F\,d\Omega$. The <photon> momentum measure is $d^3p=E^2dE\,d\Omega$, so every component of the <kinetic stress-energy tensor> has a radial factor $E^3dE=a^{-4}\epsilon^3d\epsilon$. An isotropic background has $\langle e^i\rangle_\Omega=0$ and $\langle e^ie^j\rangle_\Omega=\delta^{ij}/3$. Hence
$$
\boxed{\bar\rho=\frac{4\pi}{a^4}\int_0^\infty\epsilon^3\bar f(\epsilon)\,d\epsilon,\qquad \bar P=\frac{\bar\rho}{3}.}
$$
This <radiation pressure> equation of state requires isotropy and masslessness, not a thermal spectrum.
The distribution perturbation is $\delta f=-\epsilon\bar f'(\epsilon)\Theta$. Assume finite <energy density> and endpoint behavior $\epsilon^4\bar f(\epsilon)\to0$ at zero and infinity, as for the <Planck photon distribution>. <Integration by parts> then gives
$$
-\int_0^\infty\epsilon^4\bar f'(\epsilon)\,d\epsilon=4\int_0^\infty\epsilon^3\bar f(\epsilon)\,d\epsilon.
$$
Consequently the perturbed <energy density> and <momentum density> are
$$
\delta\rho=4\bar\rho\langle\Theta\rangle_\Omega,\qquad q^i=4\bar\rho\langle\Theta e^i\rangle_\Omega.
$$
Since $q^i=(\bar\rho+\bar P)v^i=(4\bar\rho/3)v^i$, the <photon angular temperature moments> give $\delta=4\langle\Theta\rangle_\Omega$ and $v^i=3\langle\Theta e^i\rangle_\Omega$.
The sign of <anisotropic stress> must be specified. Direct kinetic integration gives the conventional trace-free spatial stress $\pi^{ij}=T^{ij}-P\delta^{ij}=+4\bar\rho\langle\Theta(e^ie^j-\delta^{ij}/3)\rangle_\Omega$. The PDF uses the opposite sign, $\Pi^{ij}=-\pi^{ij}$, equivalently $T^{ij}=P\delta^{ij}-\Pi^{ij}$. With that convention, all requested moments are
$$
\boxed{\delta=4\langle\Theta\rangle_\Omega,\qquad v^i=3\langle\Theta e^i\rangle_\Omega,\qquad \Pi^{ij}=-4\bar\rho\left\langle\Theta\left(e^ie^j-\frac{\delta^{ij}}3\right)\right\rangle_\Omega.}
$$
Taking the spatial <trace> also gives
$$
\boxed{\delta P=\frac{\bar\rho\delta}{3}.}
$$
The angular temperature description assumes $\Theta$ is independent of <photon> energy; independent spectral distortions would require additional energy-dependent moments.
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