By no mean, this will be a holistic answer. Altough I hope it might clarify some things. In any case, any question you might have would be answered in the papers
Fluctuation-Dissipation: Response Theory in Statistical Physics (1) and the original article by Kubo (2).
1. Intro
The Fluctuation Dissipation Theorem (FDT) can be derived with a lot of different tools. Langevin equations, Fokker-Plank equations, linear response theory, phase space approach, path integral approach (MSRJD, Machlup Onsager, ... in which the FDT appears as a symmetry of the "Lagrangian").
Although it can be extended to non-equilibrium steady states (see Ref. (1)), I will only discuss about the equilibrium aspect of the theorem. Different ideas that you can get out of this theorem are that, in an equilibrium system$^1$:
a. The strength of the fluctuations is related to the dissipation in the system. For example, the Einstein relation: $D = T/\gamma$. $D$ The diffusivity, $\gamma$ a viscous damping and $T$ the temperature ($k_b=1$). Here the fluctuations cause the system to diffuse with a diffusivity $D$ which is proportional to the damping processes arising in the system.
b. Since the dissipation in the system is related to the linear response function (this is not the FDT, this is just "usual" in time invariant system). The fluctuations (noise) are related to the linear response of this system. Said differently, equilibrium thermal fluctuations are indistinguishable from a non equilibrium fluctuation produced by an infinitesimal external perturbation (which is the realm of linear response theory). This is exemplified by relations like: $\chi_A(t) = -\dfrac{1}{T}\dfrac{d\langle A(t)A(0)\rangle_{eq}}{dt}\Theta(t)$. Where $\chi_A(t)$ is the linear response function of $A$ when a constant force is applied at $t\geq0$, $\langle A(t)A(0)\rangle_{eq}$ is the autocorrelation function of $A$ at equilibrium and $\Theta(t)$ is the Heaviside function. This text in bold is very important. It is a correlation function computed at equilibrium (without external forces pushing your system). This equation tells us that the relaxation time of a thermal fluctuation (RHS of the equation) is linked to the relaxation time of the perturbation we applied (LHS of the equation, the response function)
c. Transport coefficient are more or less related to the relaxation speed of the system to an equilibrium state. Diffusivity is related to relaxation of density, viscosity to relaxation of momentum, heat conductivity of relaxation of temperature, etc... We saw above that the relaxation time from a non-equilibrium perturbation to an equilibrium state was linked to the auto-correlation of some variable at equilibrium in the linear regime. This means that we should be able to compute transport coefficients in the linear regime using well chosen auto-correlation functions. These are known as the Green-Kubo relations and are arguably the most important relations in early non-equilibrium computational physics. They read for the diffusivity: $D = \int_0^{\infty}\langle v_x(\tau)v_x(0)\rangle d\tau$. More generally, the transport coefficient for a quantity $X$ is described by an integral of the equilibrium autocorrelation of the current of $X$. For the viscosity, it would be the stress tensor autocorrelation for example.
2. Computations
This list might raise more questions than it answers. Notably, the mathematical link between all these things that we call FDT is unclear. And it is to me the major source of confusion for this topic. A way more formal treatment is given in (1) and (2), but I want to keep thing simple. We will show the link between the points a, b and c using a Brownian particle, non interacting, in the Langevin formalism. The interacting case is not really adding anything to the discussion! More generally, it might be dubious to use a Langevin equation for any statistical system. We assume that it is well justified (separation of time scales, existence of an external bath, ...)
a. The Einstein relation
Take a particle position described by a simple Langevin equation in the underdamped case with mass $m=1$:
$$\dfrac{dv}{dt}=-\gamma v + \sqrt{A}\eta(t) \text{ with } \langle \eta(t)\eta(0)\rangle=\delta(t)\text{ and } \langle \eta(t)\rangle = 0.$$
The noise is Gaussian and uncorrelated. You wrote down your equation knowing that the bath to which your particle is coupled applies a damping $\gamma$ but let's say that you don't know what should be the amplitude of the noise $A$. A thing you want, is that this particle, after a transient, reaches an equilibrium state. Which means that the velocity should sample a gaussian distribution: $$P(v) \propto e^{-\frac{v^2}{2 T}}.$$
Why should it samples this distribution? Of course because the result from our dynamical equation should agree with the results from Gibbs theory which says that the probability of observing a given state is the exponential of $H/T$. Here $H$ is $v^2/2$. Since the $v$ is driven by a Gaussian noise and is linear, $v$ will also follow a Gaussian distribution (prove it! hint: it's trivial in Fourier space!). So, we only have to chose $A$ such that the variance of the Gaussian distribution is $T$ (as required from Gibbs theory). If you know Ito theory, this is simple, otherwise you have to solve the equation. For example, by taking the Laplace transform, and turning back to real space:
$$v(t)=v_0e^{-\gamma t}+\sqrt{A}e^{-\gamma t}\int_0^{t}e^{-\gamma t'}\eta(t')dt'$$
Now, I let you compute $\lim_{t\to\infty}\langle v(t)v(t)\rangle\equiv \text{Var}(v)$ (which is not so trivial! You can ask me in the questions). Asking this variance to be equal to $T$ leads you to $A = 2\gamma T$! You have what you wanted! But how to relate that to the diffusivity to find the Einstein relation? Well, by definition the diffusivity is (with $r(0) = 0$):
$$\lim_{t\to\infty}\langle r^2(t)\rangle=2Dt.$$
Which means that you have to find $r(t)$, which is the integral of $v(t)$. You can assume, $v(0)=0$ and $r(0)=0$ without loss of generality. Then we have to compute $\langle r(t)^2\rangle$. Again, do it (you might want to look at this for some help or wikipedia on the Langevin equation). This leads to: $$\lim_{t\to\infty}\langle r(t)^2\rangle=2 \dfrac{T}{\gamma} t.$$ Which immediately gives the fluctuation dissipation relation:
$$\boxed{D = \dfrac{T}{\gamma}.}$$
This was tedious because I did it in the underdamped regime (because I will need it later). A nice exercise is to do this in the overdamped regime!
At this point, you might believe that the important part is only the Einstein relation. And of course it is, but more importantly perhaps, it the relation between the damping term and the noise term in the Langevin equation. This is the "most general" fluctuation dissipation idea. The fact that the dissipation in the Langevin equation is related to the noise term if the system is at equilibrium. Let's see why.
b. The autocorrelation part and the frequency domain. Linking damping memory to fluctuation memory.
The FDT, as it is always introduced, is way more general than what we did above. A huge assumption above was that we assumed the system to be markovian. Meaning that the state at $t + dt$ only depends on the state at $t$. However, since a langevin equation comes from the coarse-graining of a microscopic dynamics. We expect to observe some memory effect, because some degrees of freedom were "omitted" in the dynamical equation but should still affect the system. Thus, it makes sense to restate the equation as a Generalized Langevin Equation (GLE):
$$\dfrac{dv(t)}{dt}=-\int_{-\infty}^{t}\gamma(t'-t)v(t')dt' + \eta(t).$$
If you imagine that this Langevin equation describe the velocity of a big particle surrounded by a sea of small ones. You can pretty much say that $\eta$ is the effective random noise produce by collisions between the small particles and the lage one while $\gamma$ is an effective damping, due to these collisions. However, the displacement of the big particle, will surely cause the small particles to react in some way. For example, if you consider them as a fluid, the big particles will impart some velocity to this fluid and you have somehow to keep track of this velocity for the surrounding fluid. Here you have two choices, either you make the velocity of the fluid an additional degree of freedom (in which case you don't need any memory), or you do as we did just above and include this additional degree of freedom as some sort of memory. Here in $\gamma(t-t')$. If the relaxation time of the integrated degrees of freedom is of order $\tau$, then it is a great assumption to simply guess $\gamma(t-t') = -\gamma_0e^{-|t-t'|/\tau}$.
Anyway. Again, we require consistency with equilibrium ($\langle v^2\rangle=T$). You can show that, for this to be the case, the noise too must have time correlation (memory!). How to see that and more importantly, what should be the correlations of the noise quantitatively in order for this GLE to describe an equilibrium system? First let's find the relation between the autocorrelation and the response. We conveniently define $\Gamma(t) = \gamma(t)\Theta(-t)$, (the system cannot have memory of the future) which allows us to rewrite the GLE as (see the boundaries of the integral):
$$\dfrac{dv(t)}{dt}=-\int_{-\infty}^{\infty}\Gamma(t'-t)v(t')dt' + \eta(t).$$
Take the Fourier transform of both sides:
$$-iw v(w)=-\Gamma(w)v(w)+\eta(w)\Rightarrow v(w)=\dfrac{\eta(w)}{\Gamma(w)-iw}$$
The response function is $\chi = (\Gamma(w) - iw)^{-1}$ (prove that it is indeed a linear response function!). We want to relate that to the autocorrelation function. To do this, let's rewrite a bit differently the GLE:
$$\dot v(t_0+t)+\int_{t_0}^{t_0+t}\gamma(t_0+t-t')v(t')dt'=-\int_{-\infty}^{t_0}\gamma(t_0+t-t')v(t')dt'+\eta(t_0+t)$$
The usefulness of this decomposition comes from the fact that the
RHS is uncorrelated to the velocity at $t_0$ (see Fluctuation-dissipation theorems from the generalised Langevin equation). Therefore we can average with $v(t_0)$ and get rid of the RHS:
$$\langle v(t_0)\dot v(t_0+t)\rangle+\int_{t_0}^{t_0+t}\gamma(t_0+t-t')\langle v(t')v(t_0)\rangle dt'=0.$$
Now, take a "fourier like" transform:
\begin{split}
0&=\int_0^{\infty}\langle v(t_0)\dot v(t_0+t)\rangle e^{iwt}dt+\int_0^{\infty} \int_{t_0}^{t_0+t}\gamma(t_0+t-t')\langle v(t')v(t_0)\rangle e^{iwt}dt'dt\\
&=-iw\int_0^{\infty}\langle v(t_0) v(t_0+t)\rangle e^{iwt}dt + \langle v(t_0)^2\rangle+\Gamma(w)\int_{0}^{\infty}\langle v(0)v(t)\rangle e^{iwt}dt
\end{split}
(I let you do the math, it's is not trivial. You have to take care of the boundaries in the integrals and assume time translation invariance). It gives (using again time translation invariance):
$$\boxed{\chi(w) = -\dfrac{1}{\langle v^2\rangle}\int_0^{\infty}\langle v(t)v(0) \rangle e^{iwt} dt}$$
Using $\langle v^2\rangle = T$, we obtain the relation I gave in the intro (in $w$ space, without a derivative because it depends on the variable we chose.. Velocity or position.). This relates the linear response to the autocorrelation.
I promised you that we were going to relate the memory of the damping with the one of the noise (which is really a fluctuation dissipation theorem because it links the memory of the noise (fluctuation) with the memory of the damping (dissipation)), so here we go.
The autocorrelation of $v$, as we already saw, depends only on the time difference due to time translation invariance: $$\langle v(t')v(t)\rangle = \langle v(t'-t)v(0)\rangle.$$ Actually we can apply something stronger: $$\langle v(t'-t)v(0)\rangle=\langle v(|t'-t|)v(0)\rangle$$
The autocorrelation is even! Which tells us that its fourier transform verifies:
$$\int_{-\infty}^{\infty} \langle v(t)v(0)\rangle e^{iwt}dt=2\Re\left\{\int_0^{\infty}\langle v(t)v(0)\rangle e^{iwt}dt\right\}=2T\Re\{\chi(w)\}$$
Where $\Re$ is the real part (prove the first equality from the evenness of the autocorrelation!). The last equality follows from the last boxed equation. Note that I corrected a minus sign. I don't know where the error is coming from, it must be a stupid convention in the Fourier transform that I miss :) (but it's not too important). To go further we need to express the fourier transform autocorrelation of the velocity as some kind of thing related to the noise.
We recall the relation from above:
$$v(w)=\dfrac{\eta(w)}{\Gamma(w)-iw}\Rightarrow \langle v(w)v(-w) \rangle = \langle \eta(w)\eta(-w)\rangle |\chi(w)|^2$$
We can make use of a very useful theorem. The Wiener-Kinchin theorem that relates power spectrum and autocorrelation:
$$\int_{-\infty}^{\infty}\langle X(0)X(t)\rangle e^{iwt}dt = \langle X(w)X(-w)\rangle.$$
(prove it!). Applied to the equation above, this immediately leads to:
$$2T\Re\{\chi(w)\} = \langle \eta(w)\eta(-w)\rangle |\chi(w)|^2.$$
Moreover, we know that $\chi(w) = (iw-\Gamma(w))^-1$ which implies that:
$$\boxed{2T\mathcal{\Re}\{\Gamma(w)\}=\langle \eta(w)\eta(-w)\rangle}$$
Which is what we wanted! A relation linking the fluctuation (RHS) to the dissipation (LHS). Note that here, we have the full temporal/frequency dependence. From this equation we get that, at equilibrium, it is required that the autocorrelation of the noise decays as the memory kernel $\gamma(t)$ of the damping. A pretty strong result indeed!
Note that here, I really played with different notation, fourier transform (look at all the possible boundaries), the heaviside functions, ... A lot of things have been put under the carpet, notably concerning causality (which often requires some Heaviside function, or some change of the boundaries from $-\infty$ to 0, ..). Thus, It's likely that if you try to perform the same computations on your side, you obtain different results. The first check for these types of computations should be related to the boundaries of the integral and the causality.
Two nice exercises are: 1. to rederive the section a. by assuming a delta correlated noise and 2. to derive the FDT for the position when an additional linear force is added (see https://arxiv.org/abs/cond-mat/0504750 if you struggle. It's also a very nice learning resource taking a different route (easier) than mine)
c. Green-Kubo relations.
I'm getting tired of writing so I'll make it quick. Let's get back to the expression of the mean square displacement from the integration of the velocities:
\begin{split}
\langle r^2\rangle&=\left\langle \int_0^{t}\int_0^{t}v(t')v(t'')dt'dt''\right\rangle\\
&=2\int_0^{t}(t-\tau)\langle v(0)v(\tau)\rangle d\tau
\end{split}
Where the second equality follows from time translation invariance (prove it!).
Now, we take back our relation from earlier: $D=\lim_{t\to\infty}\langle r^2(t)\rangle/(2t)$. If the autocorrelation decays fastly (this is a problem in fluctuating hydrodynamics), we obtain:
$$\boxed{D = \int_0^{\infty}\langle v(0)v(t)\rangle dt}$$
Which is the Green-Kubo relation for the diffusivity. A very important fluctuation dissipation theorem. Here it is a very precise example, but as said in the intro, you can write this relation for any current you want, and you will obtain the transport coefficient associated to it. It's very useful, because it relates macroscopic quantities (transport coefficient of macroscopic field) to microscopic quantities (microscopic current of the transported quantity).
A last thing, prove from this relation and one of the (boxed) relation above that transport coefficients are related to the integral of response functions! An other FDT!
2. Outro
As you saw, it's difficult to really say what's the FDT because it comes with different flavors. But the most important ones are the ones discussed in b. because they are the most general out of which every thing can be derived.
$^1$ Equilibrium system $\to$ A system that is Hamiltonian and with a steady state sampling a canonical distribution for example.
