I don't understand how the resistor between GND and the source in a common-source amplifier stabilizes it
How does Rs stabilize the amplifier?
\$V_{GS} = V_G - V_S \$
\$V_S = I_D R_S \$
\$\Rightarrow V_{GS} = V_G - I_D R_S \$
Now, \$V_G\$ is constant so, if \$I_D\$ increases, \$V_{GS}\$ decreases.
But, if \$V_{GS}\$ decreases, \$I_D\$ decreases.\$^1\$
Thus, an increase in \$I_D\$ acts to decrease \$I_D\$. This is the hallmark of negative feedback.
\$^1 I_D \approx K(V_{GS}-V_{TH})^2 \$
The gate to source voltage, \$V_{GS}\$, controls the transistor. As this voltage increases, the channel becomes more conductive, and more current flows through the drain and source.
The voltage across a resistor is given by Ohm's law:
$$ v = I R $$
So, as more current flows the drain and the source, more current flows through \$R_S\$, and, by Ohm's law, the voltage across \$R_S\$ increases. Since the gate voltage isn't changing, the gate-source voltage decreases, since the source is now closer to the gate.
There will be one point at which this is stable. If the transistor is too much off, there won't be enough current in \$R_S\$, and \$V_{GS}\$ will be high enough to turn the transistor on more, increasing the current in \$R_S\$.
If the transistor is too much on, there will be too much current in \$R_S\$, and \$V_{GS}\$ will be low enough to turn the transistor off more, decreasing the current in \$R_S\$.
If the gain of the transistor is infinite, then as long as the input is not pushing the output into the supply rails, then the gate-source voltage will be constant, right at the transistor's threshold voltage: \$V_{GS} = V_{th}\$. Since the transistor's gain is infinite, it has unlimited ability to correct any deviation from this through the feedback mechanism mentioned.
The gain of the circuit approaches \$R_D/R_S\$ as the transitor's gain increases. Or put another way, as \$R_D/R_S\$ decreases, the transistor's gain becomes less relevant to the gain of the whole circuit. This is how the simplifications you mention are made. That is, almost all of the change in gate voltage appears as a change in \$V_{R_S}\$, and only a negligible amount as a change in \$V_{GS}\$.