Vds meaning in characteristic curve of a MOSFET
As far as I understand, drain-source voltage means &quot;while MOSFET doesn't conduct&quot;

No, the drain-source voltage \$V_{DS}\$ applies too all situations, including the situation where the MOSFET conducts.
Maybe you're confused because in these curves:

\$V_{DS}\$ can be quite large, like 2 Volts or more.
Indeed then the MOSFET isn't fully conducting, note how the curve for \$V_{GS}\$= -2.5 V at \$V_{DS}\$ = 2 V results in \$I_D\$ = 5 A meaning the MOSFET's resistance (between drain and source) is about 0.4 ohms. That's much higher than the \$R_{DS,on}\$ we generally would want to have (like the 50 mOhm you mention) when switching things.
The difference here is that \$V_{GS}\$= -2.5, meaning the MOSFET isn't fully switched on. To fully switch it on, we need a higher \$V_{GS}\$. See how the current increases a lot as \$V_{GS}\$ is increased to -5 V or -10 V. Then the MOSFET is fully turned on and it will have a low \$R_{DS,on}\$.
\$V_{DS}\$ is always simply the voltage across drain-source, independent of the mode of operation of the MOSFET (conducting or not). The curves show us that we need to apply a large \$V_{GS}\$ in order to keep \$V_{DS}\$ low for large currents. But suppose I only want to switch a small current and/or I don't care that \$R_{DS,on}\$ isn't as low as it could be. Maybe I want to apply a lower \$V_{DS}\$ because my circuit doesn't have 5 V available. Then I can see what voltage drop (\$V_{DS}\$) I get for a certain current.