To your first point of "are they one-body or many-body ones?", I am not quite sure what you mean. This quantity involves all of the many-body eigenstates, but it is a one-body quantity in the sense that it only accounts for processes of adding/removing one particle.
From the definition of the retarded Green's functions we have
$$ G^{r} = -i\Theta(t-t')\frac{1}{Z}\text{Tr}\Big[e^{-\beta(\hat{H}-\mu \hat{N})}\{\hat{\psi}(xt),\hat{\psi}^{\dagger}(x't') \} \Big] $$
where $\{\cdot,\cdot\}$ is the anti-commutator and $ Z = \text{Tr}\exp(-\beta(\hat{H}-\mu\hat{N}))$, and the trace is over all the many-body eigenstates. The advanced is defined analogously:
$$ G^{a} = i\Theta(t'-t)\frac{1}{Z}\text{Tr}\Big[e^{-\beta(\hat{H}-\mu \hat{N})}\{\hat{\psi}(xt),\hat{\psi}^{\dagger}(x't') \} \Big] $$
Assuming a time-independent $\hat{H}$, we fourier transform to $t-t' \to \omega$, from the Lehmann representation we have that:
$$ G^{r}(\omega) - G^{a}(\omega) = 2i\ \text{Im}G^{r}(\omega) $$
Substituting this relation back into your expression gives,
$$ G^{<}(\omega) = - 2if(\omega)\text{Im}G^{r}(\omega), $$
which shows a much more immediate analogy to your standard fluctuation-dissipation relation. That is, an analog that connects a correlation function and the imaginary part of a response function. I am unaware of a deeper physical intuition in this expression, so if someone else has a better sense, please feel free to correct my answer.
A great source is this chapter submitted to the Autumn School on Correlated Electrons: https://arxiv.org/abs/1907.11302