Fluctuation dissipation on a ring? The integral fluctuation theorem is given by:
$$\left< e^{-R}\right>=1\tag{0}$$
where:
$$R\equiv \ln \left( \frac{p_0(\vec n_0) p[\vec n(\tau),\vec c(\tau)]}{p_f(\vec n) \cdot p[\tilde n(\tau),\tilde c(\tau)]}\right)\tag{1}$$
where my notation follows in part (arxiv:0605080) with $p[\vec n(\tau),\vec c(\tau)]$ being the trajectory weight for a given trajectory $\vec n(\tau)$ with initial state $\vec n_0$. And $ p[\tilde n(\tau),\tilde c(\tau)]$ being that for the trajectory $\tilde n(\tau)\equiv n(t-\tau)$ under the time reversed protocol $\tilde c(\tau) \equiv c(t-\tau)$. Lastly $p_0(\vec n_0)$ is the initial distribution and $p_f(\vec n)$ the finial distribution.
From this on a ring it is possible to derive the equation:
$$p(-\Delta s_{tot})=e^{-\Delta s_{tot}}p(\Delta S_{tot})\tag{2}$$
where $p(\Delta S_{tot})$ is the pdf for total entropy production. Following back references it appears that this relation originated in (Crooks, 1999).
My question: I understand how in general such relations are derived but I can't see where the $p(\Delta s_{tot})$'s in equation (2) come from in relation to equation (1). I.e. how do we relate $p(\Delta s_{tot})$ to probabilities in equation (1)?
 A: Short answer
The Crooks relation cannot be derived (as far as I can tell) directly from the "integral fluctuation theorem" (eq. (0) in the OP) and a generalized theorem is required - which the rest of this answer will look into.
Notation
I will follow the notation in the reference {1} below. This is summarized here:

*

*$\dagger$ denotes quantities relating to the time-reversed process.

*$S_\alpha[x(\tau)]$ is a functional of the original dynamics and:
$$ S_\alpha^\dagger[x(\tau)^\dagger, \lambda^\dagger, F^\dagger]=\varepsilon_\alpha S_\alpha[x(\tau), \lambda, F]$$

*$g$ is a function depending on an arbitrary number of functionals $S_\alpha$
Generalized Theorem
The generalized theorem is then given by:
$$\langle g(\{ \varepsilon_\alpha S^\dagger_\alpha[x^\dagger(\tau)] \})\rangle^\dagger=\langle g(\{ \varepsilon_\alpha S_\alpha[x(\tau)] \}) \exp(-R[x(\tau)])\rangle \tag{A1}$$
This is proved on page 7 of {1} and as such I will not reproduce the proof here.
Deriving Crooks theorem
Both reference {1} and {2} then goes into explaining how we derive crooks theorem. Let us first consider $R[x(\tau)]$. We start both the original and reversed dynamics in the stationary state. This means that:
$$R=\frac{\Delta S_m}{k_B}+\ln \left( \frac{p_i(\vec x_0)}{ p_f(\vec x)}\right)$$
$$=\frac{\Delta S_m}{T}+\frac{ \Delta V-\Delta \mathcal{F}}{T}$$
where $\Delta V$ is the change in potential and $\Delta \mathcal{F}$ the change in free energy. Using
$$W[x(\tau)]=\Delta S_m+ \Delta V$$
we get:
$$R=(W[x(\tau)]-\Delta \mathcal{F})/T$$
Further more we chose that $S_\alpha[x(\tau)]=W[x(\tau)]$ (which corresponds to $\varepsilon_\alpha=-1$) and take:
$$ g(W[x(\tau)])=\exp(-k W[x(\tau)])$$
Thus (A1) becomes:
$$\langle \exp(k W[x(\tau)]) \rangle^\dagger =\exp(\Delta \mathcal{F}/T)\langle \exp(-k W[x(\tau)]-W[x(\tau)]/T) \rangle \tag{A2}$$
Note that:
$$\langle \exp(-\alpha W[x(\tau)]) \rangle=\int P(W)e^{-\alpha W} dW$$
so taking the inverse Laplace transform of (2) w.r.t $k$ gives us:
$$P^\dagger(-W)=P(W) \exp(\Delta \mathcal{F}/T-W[x(\tau)]/T)$$
which is the Crooks theorem as given in terms of work rather then entropy as given in the OP.
References

*

*Stochastic Thermodynamics by Luka Pusovnik. Available from: https://mafija.fmf.uni-lj.si/seminar/files/2016_2017/luka-pusovnik-stochastic-thermodynamics.pdf


*Täuber, U.C., 2014. Critical dynamics: a field theory approach to equilibrium and non-equilibrium scaling behavior. Cambridge University Press. (pg 338)


*See also https://arxiv.org/abs/1201.6381
