= Solution
Use the <Collisionless Boltzmann equation> in <action-angle variables>,
$$
\partial_tf+\{f,H\}=0,\qquad \{f,H\}=\partial_{\boldsymbol\theta}f\cdot\partial_{\mathbf J}H-\partial_{\mathbf J}f\cdot\partial_{\boldsymbol\theta}H.
$$
Define the orbital <frequency> vector by
$$
\boxed{\boldsymbol\Omega(\mathbf J)=\partial_{\mathbf J}H_0=(\Omega_\varphi,\Omega_R).}
$$
The slow evolution of the angle average $f_0$ starts at second order in the perturbation. At first order, the <Poisson bracket> therefore gives
$$
(\partial_t+\boldsymbol\Omega\cdot\partial_{\boldsymbol\theta})\delta f=\partial_{\mathbf J}f_0\cdot\partial_{\boldsymbol\theta}\delta\Phi.
$$
For the $\mathbf k$th term in the <Fourier series>, this reads
$$
(\partial_t+i\mathbf k\cdot\boldsymbol\Omega)\delta f_{\mathbf k}=i\mathbf k\cdot\partial_{\mathbf J}f_0\,\delta\phi_{\mathbf k}.
$$
The lowercase $\delta\phi_{\mathbf k}$ here denotes the <Fourier coefficient> of the uppercase $\delta\Phi$ in the <Hamiltonian>; these are the same perturbing potential. Multiply by the <integrating factor> $e^{i\mathbf k\cdot\boldsymbol\Omega t}$ and set the initial homogeneous response to zero. One obtains
$$
\boxed{\delta f_{\mathbf k}(\mathbf J,t)=i\int_0^t dt'\,\mathbf k\cdot\frac{\partial f_0(\mathbf J,t')}{\partial\mathbf J}\,e^{-i\mathbf k\cdot\boldsymbol\Omega(t-t')}\delta\phi_{\mathbf k}(\mathbf J,t').}
$$
Retaining the slow time dependence of $f_0$ inside this <integral> is compatible with the subsequent second-order calculation; changes in $\boldsymbol\Omega$ caused by that slow evolution affect only higher orders in this approximation.
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