= Solution
Let $\langle\cdot\rangle$ denote $(2\pi)^{-2}\int d^2\theta$. Average the <Collisionless Boltzmann equation> to second order. All first-order angle derivatives average to zero, leaving
$$
\partial_tf_0=-\langle\{\delta f,\delta\Phi\}\rangle.
$$
An angle <integration by parts> shows that
$$
\begin{aligned}
\langle\{\delta f,\delta\Phi\}\rangle
&=\left\langle\partial_{\theta_i}\delta f\,\partial_{J_i}\delta\Phi-\partial_{J_i}\delta f\,\partial_{\theta_i}\delta\Phi\right\rangle\\
&=-\partial_{J_i}\langle\delta f\,\partial_{\theta_i}\delta\Phi\rangle.
\end{aligned}
$$
Thus the angle-averaged <galactic distribution function> obeys an action-space <continuity equation>,
$$
\partial_tf_0=-\partial_{J_i}q_i,\qquad q_i=-\langle\delta f\,\partial_{\theta_i}\delta\Phi\rangle.
$$
This sign also follows from <Hamilton's equations>, $\dot J_i=-\partial_{\theta_i}\delta\Phi$: the flux is $\langle\delta f\,\dot J_i\rangle$.
Reality gives $\delta\phi_{-\mathbf k}=\delta\phi_{\mathbf k}^*$ and $\delta f_{-\mathbf k}=\delta f_{\mathbf k}^*$. Orthogonality of the <Fourier series> implies
$$
\langle\delta f\,\partial_{\theta_i}\delta\Phi\rangle=-i\sum_{\mathbf k}k_i\delta f_{\mathbf k}\delta\phi_{\mathbf k}^*.
$$
Insert <3/b> to obtain
$$
q_i=-\sum_{\mathbf k}k_i\int_0^t dt'\,\mathbf k\cdot\partial_{\mathbf J}f_0(\mathbf J,t')\,e^{-i\mathbf k\cdot\boldsymbol\Omega(t-t')}\delta\phi_{\mathbf k}(\mathbf J,t')\delta\phi_{\mathbf k}^*(\mathbf J,t).
$$
Although individual terms are complex, pairing $\mathbf k$ with $-\mathbf k$ makes the flux real.
Now take the <marginalization> specified by $F_0=2\pi\int_0^\infty f_0\,dJ_R$. Assume vanishing radial-action flux at $J_R=0$ and $J_R\to\infty$: the first boundary does not permit negative <radial action>, and the second has no escaping flux in this calculation. Then
$$
\partial_tF_0=-2\pi\partial_{J_\varphi}\int_0^\infty q_\varphi\,dJ_R-2\pi[q_R]_0^\infty=-\partial_{J_\varphi}Q_0,
$$
with
$$
\boxed{\begin{aligned}
Q_0={}&-2\pi\sum_{\mathbf k}k_\varphi\int_0^\infty dJ_R\int_0^t dt'\,\mathbf k\cdot\frac{\partial f_0(\mathbf J,t')}{\partial\mathbf J}\\
&\quad\times e^{-i\mathbf k\cdot\boldsymbol\Omega(t-t')}\delta\phi_{\mathbf k}(\mathbf J,t')\delta\phi_{\mathbf k}^*(\mathbf J,t).
\end{aligned}}
$$
If radial-action boundary flux were present, its explicit boundary term would have to be retained.
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