Since you find the principles of TD-OCT much easier to understand, just before the spectrometer, SD-OCT is exactly the same as TD. Inside the spectrometer diffraction from the grating is a Fourier Transform (FFT)! , so you can assume that the broadband light divided into its wavelengths (definition of FFT). similar for the SS-OCT, where you have all the wavelengths and then integral all of those (IFFT) in the single detector.

Let me try.

If you understand TD-OCT, it is great. So you understand that the wider the bandwidth of the light source, the shorter the coherence length (higher resolution of OCT image). That means if you use infinity bandwidth, you will have infinitely short coherence length and the signal would be only exactly at dz=0 (d_obj = d_ref). So now, let consider in other way. Start limit your bandwidth. You shall find that signal oscillation does not disappear so rapidly as before, so there is some vicinity of positions around dz=0 at which you still can see the signal. If you reduce bandwidth more and more up to exactly single wavelength you find that no matter how far you are from condition dz=0, there is always high signal (it varies as cosine signal due to interference of ref and obj light while moving the ref mirror). Ok then, so let detect this single wavelength signal by using a spectrometer in SD-OCT. What you will see is a single pixel which amplitude varies as cosine while moving reference mirror. Nothing special. So let's include a second wavelength in your light source so now you have a two-wavelength laser. Let the second wavelength differ not so much. Consider SD-OCT with a spectrometer having field of view of 100nm and having 2048 pixels operating at 1050nm center wavelength. So the second wavelength would be 1000nm + 100nm/2048 = 1000,05nm. How the signal will look like? Signal at first pixel still follows cosine cos[k(1000nm)*dz(t)]. The signal at the second pixel also shows a cosine according to cos[k(1000,05nm)*dz(t)]. As you see the phase (argument of cosine) would be slightly different. Assume dz=1mm then phase of the first pixel k1*dz=6283.2 and the second pixel k2*dz=6282.9, so it differs by 0.3 rad (17 deg), correct? You can check that at the 2048th pixel (1100nm) the phase is 5712. Now just calculate phase and cosine(phase) for each of 2048 pixel in excel and make a graph. You will see nice cosine fringe spanning along the pixels - this is your virtual spectrometer. Now put very small dz like 10um. You would see a fringe pattern which has just a single period along all pixels. At dz=0 all phases are zero and cosine(0)=1 for all wavelengths. That means in SD-OCT you see no signal! This is curious because in TD-OCT you observe maximal signal at exactly dz=0, right? Yes. Because in TD-OCT you observe a sum of all electric fields of all wavelengths by using a single photodiode. That means peaks and valleys of electric fields must match each other to produce highest interference signal. Since the peaks and valleys of electric field of all wavelenghts match perfectly (are highly coherent), they produce highest interference signal exactly at dz=0.

For dz > 0 phases of all wavelengths differs too much and corresponding electric fields rapidly loose the coherence, so the sum of them quickly decay to ~zero. As you saw in above virtual spectrometer considerations, at dz=1mm a two very close wavelengths (1000 and 1000.05) differs already by 17 deg. That means a wavelength in 10th pixel is at phase 170deg vs 1st pixel, at 20th pixel it is ~360deg. So you have one oscillation at 20 pixels, ~100 oscillations at 2048 pixels. At spectrometer you still observe nice interference fringes. But if you sum all of them (to simulate signal at TD-OCT - this is how TD-OCT observe signal) you will find that signal is zero! Sum of cosine signal along 100 oscillations would always give ~0 value. But if you observe a signal at spectrometer which has less than one oscillation, sum of it is > 0 and is highest when all pixels have 1.0 :). This is the all magic of TD and SD OCT :).

One digression. In TD-OCT if you don't move the ref mirror, signal from a photodiode is constant in time because neither k nor dz don't depend on time. To observe a signal you need to make k(t) or dz(t). In TD-OCT you can only make dz(t) by moving ref mirror so the signal at photodiode start changing oscillating with cost(k*dz(t)). Please note that you may also make k(t) - this is Swept-Source OCT where the wavelength of the swept laser is changing over the time. In SS-OCT you also use a single photodiode but you don't observe all wavelengths at once anymore like in TD-OCT. So SS-OCT is just "time-spectrometer" OCT :). I hope you will have better understanding now.

So finally how the depth is encoded in SD-OCT?

In SD-OCT every surface at certain depth creates its own signal of certain frequency. For example let have a two surface object, one surface is at dz=1mm and second one at dz=2mm. First surface produce frequency which has 1 oscillation per 20 pixels (as discussed above), second surface will produce 1 oscillation per 10 pixels. So you see two frequencies at your spectrometer. Then you need to use Fourier Transform to see these frequencies as two peaks.

By the way, now you may see that there would be limited dz=dz_max which you can observe. It is such dz_max which produce one oscillation per 2 pixels at the spectrometer. For above example it is ca. 5.6mm. Above the dz_max you will observe aliased signals. But this is another story.