The standard coordinate system is the mathematically simplest, but I don't think it's actually the most physically intuitive. This is because we live on objects that are gravitationally bound, and admist objects that are electromagnetically defined. This means that our local length scales are not affected by cosmological expansion. But, if you look at the FRLW metric, in its standard form (I choose the flat cosmology for simplicity):

$$ds^{2} = - dt^{2} + a(t)^{2}\left(dr^{2} + r^{2}d\theta^{2} + r^{2}\sin^{2}\theta d\phi^{2}\right)$$

you can tell that, for some constant-t observer, the ruler actually expands with time by a factor $a$. For this reason, when describing cosmological observations, I actually like to use a different coordinate system, where you replace $r$ with $R = a(t)r$. Then, you have $dR = {\dot a}r\,dt + a\,dr \rightarrow dr =dR- H (R/a)\,dt$, and the metric becomes (note that I used the relation $H = \frac{\dot a}{a}$, to replace $a$ with Hubble's "constant"):

$$ds^{2} = -\left(1-H^{2}R^{2}\right)dt^{2} + 2dR\,dt\left(-HR\right) + dr^{2} + r^{2}d\theta^{2} + r^{2}\sin^{2}\theta d\phi^{2}$$

In terms of direct physics, this coordinate system is a lot clearer. You see that, for a constant-t observer, there is a coordinate singularity at $R = \frac{1}{H}$, corresponding to the cosmological horizon. Furthermore, this coordinate system has a $g_{tr}$ coordinate, which, it can be shown, corresponds to the frame dragging of the system -- so space naturally expands away at a velocity proportional to $HR$, which gives you Hubble's law.

The standard coordinate system is the mathematically simplest, but I don't think it's actually the most physically intuitive. This is because we live on objects that are gravitationally bound, and admist objects that are electromagnetically defined. This means that our local length scales are not affected by cosmological expansion. But, if you look at the FRLW metric, in its standard form (I choose the flat cosmology for simplicity):

$$ds^{2} = - dt^{2} + a(t)^{2}\left(dr^{2} + r^{2}d\theta^{2} + r^{2}\sin^{2}\theta d\phi^{2}\right)$$

you can tell that, for some constant-t observer, the ruler actually expands with time by a factor $a$. For this reason, when describing cosmological observations, I actually like to use a different coordinate system, where you replace $r$ with $R = a(t)r$. Then, you have $dR = {\dot a}r\,dt + a\,dr \rightarrow dr =dR- H (R/a)\,dt$, and the metric becomes (note that I used the relation $H = \frac{\dot a}{a}$, to replace $a$ with Hubble's "constant"):

$$ds^{2} = -\left(1-H^{2}R^{2}\right)dt^{2} + 2dR\,dt\left(-HR\right) + dR^{2} + R^{2}d\theta^{2} + R^{2}\sin^{2}\theta d\phi^{2}$$

In terms of direct physics, this coordinate system is a lot clearer. You see that, for a constant-t observer, there is a coordinate singularity at $R = \frac{1}{H}$, corresponding to the cosmological horizon. Furthermore, this coordinate system has a $g_{tr}$ coordinate, which, it can be shown, corresponds to the frame dragging of the system -- so space naturally expands away at a velocity proportional to $HR$, which gives you Hubble's law.