They're not really moving faster than light, and there's no objective sense in which "expanding space" is making them move faster than light, either. (Expanding space is just a convention and not anything physical.) See physics.stackexchange.com/q/400457/180843

so the rate of expansion between two objects beyond a certain distance cannot be not faster than light?

There is a difference between "the object moves" and "the space expands".

Does this answer your question? Why speed of light is considered to be the fastest?

Is the question "how are cosmological recession rates able to exceed the speed of light?" (should be answered in the question @hobbs linked, but I'm sure you saw it already, given that you asked it), or is the question "what is the physical process that gave things these high recession rates?"

@Sten, I guess it's a combination of both in a single question. I am thinking like what propels or makes the object go/move away that fast. Please note that I am not a scientist. I am just an inquisitive individual so a simple yet thorough explanation will really help.

@hobbs, yes the answer of that question did help but I am more interested in knowing how an object gets to move away faster than light.

If you read that question Ed, you'll see that it is not moving away faster than light. There is not a simple answer, unless you understand the maths involved.

The typical metaphor is two ants standing on a balloon that you're blowing up. One ant is walking away from the other at less than the speed of light, but the balloon itself is expanding so the space between the ants is increasing in a way that has nothing to do with the movement of the ant.

@Sten Is Hubble's law not an objective sense? Does it not make sense to say that, right now, the distance to GN-z11 is 32 Glyr but now, one second later, it's 660,000 km more, and having increased its distance by 660,000 km in one second, it has receded by 2.2 times the speed of light? (sorry for perhaps repeating myself, but I just don't get why it doesn't make sense to say that it recedes faster than the speed of light).

@pela The distance along successive comoving synchronous hypersurfaces, with respect to the clock times of comoving observers on those surfaces, does indeed grow at that rate. But there's no reason we had to pick those definitions of distance and time. They are only convenient because they appropriately respect the cosmological principle.

@pela But for example, you could use the same definitions to conclude that the separation between Earth and a spaceship departing at close to $c$ is increasing at a superluminal rate (see point #2 of this answer). Cosmic expansion isn't special in this respect.

@Sten If I understand that example correctly, it's equivalent to saying "If I travel 1 lightyear at 99% the speed of light, so that $\gamma\simeq7$, then in my reference frame it will only take 1/7 year, so I must have traveled at 7 times the speed of light". Is that what you're saying?

@pela Yeah, exactly. The main complication beyond that is to make two observers (e.g. earth and spaceship) symmetric, to correctly correspond to how we treat comoving observers. That's the reason for the $\pm\vec v$ velocities in the example. But it's still exactly the time dilation that lets the recession speeds be superluminal.

@Sten That seems to me a bit like cheating. The 7c speed in my example above is calculated from using the traveler's time, but the stationary observer's measure of distance. How can you justify not using the traveler's measure of distance, which is only ~1/7 lightyears?

@pela It is cheating -- that's why we shouldn't worry that the speed of light is exceeded! :) But it corresponds pretty closely with cosmological recession rates. Roughly speaking, distant objects are receding so fast because we are using our distance and their clock. (Just a rough picture, really we're picking coordinates that transition smoothly from our frame to theirs.)

@Sten more to the point, they are not "measured" to be moving faster than the speed of light either (what would the redshift of something be travelling at $>c$) ?

@Sten But… but… When we talk about the recession velocity of a distant galaxy, we mean the rate at which it increases its distance. That distance is "our" distance (i.e. physical, i.e. "let's freeze the Universe and lay out meter sticks"), I agree. But the "rate" is given by our clock, not theirs, no? I mean, we define $v_\mathrm{rec}$ as [how many kilometers does our "measured" (i.e. inferred) distance increase] divided by [how many seconds did our clock advance].

@pela The FLRW spatial surfaces (what I called "comoving synchronous" above) are the surfaces on which comoving observers agree on how much time has elapsed since the Big Bang. That should sound immediately suspicious, because when we think of e.g. a spaceship departing from Earth at close to the speed of light, we don't think of it agreeing with the Earth on how much time has elapsed! But those are the surfaces on which the distances are conventionally measured, and the cosmological time conventionally corresponds to the clocks of the comoving observers on such a surface.

Here's a visual; black lines are comoving observers while gray curves are spatial surfaces. So regarding your question, the "rate" is given by both our clock and theirs. This is why benrg's answer (linked above) is helpful, because it gives an example within a frame where the two observers' clocks are synchronous. That's almost exactly what's happening with cosmological recession (at least in a flat spacetime).

The only difference is that the distance in benrg's answer would be measured along a straight line in the diagram rather than following the curvature of the appropriate gray spatial surface.