Id / Vgs is certainly a ratio appears to have no significance, just as a diode V/I has no significance as fixed ratio where we call the derivative ESR or Rs.
But let me show you my insight.
We would expect the 1st derivative of this function is quite different than the function. 1st yr calculus right? This derivative can be measured as \$\frac{\Delta Id}{\Delta V_{gs}}=g_m\$
When you choose to operate in the linear mode at a set bias Id, then you choose the Rd , compute gm and compute the Drain voltage gain.
But since this curve is exponential and we know the derivative of d(e^x)/dx is the same e^x dx, and derivative of \$e^{kx}~~ \text{is just }~~ke^{kx} dx\$
Thus k can be seen in the graph I made below from the blue curve by a constant gap or ratio.
What is the difference between upper equation and gm = Id/Ugs (without changes)?
Note that k appears to be constant (30+/-1) up to the Vgs threshold of 4V. AtThe difference is a ratio k which appears to be constant (30+/-1) up to the Vgs threshold of 4V.
At this point both gm are flattening out as RdsOn rapidly reduces to its rated
ON resistance and gm has very low gain.
Is it relevant? not really. Unless you find some use for it.
Ideally one should use 3x Vth minimum or Vgs =12V to get near rated RdsOn and use Vgs < Vth=4V for higher gm values. However the resistance is higher so there are other factors if one wants to choose a linear operating point for max power gain.