How is a four terminal sensing compensating conductor resistances?
why are these conduction resistances negligible?


Because a voltmeter ideally has infinite resistance, and in practice is typically 10 megohms or more.

In the case of the current meter that means the current it measures is very nearly the same as the current through the device under test (DUT).

And in the case of the voltmeter it means the current through the leads that connect it to the DUT is very small, and therefore any resistive voltage drops in those leads is also very small.

Edit 

In comments you said


  but I still have voltage drops in the conductions from the DUT to the voltmeter?!


Let's work an example. Suppose you're trying to measure the resistance of a 1 ohm resistor, and your interconnect wires have 10 milliohms resistance each. We'll use a 10 mA probe current. So the two measurement scenarios look  like this:







simulate this circuit &ndash; Schematic created using CircuitLab

Let's assume our voltmeter has 1 megohm burden resistance.
Now in the 2-wire measurement, the ammeter reads 10 mA and the voltmeter reads .010199 V. So we estimate the resistance of our DUT to be about 1.02 ohms.

In the 4-wire measurement, the ammeter again reads 10 mA (being in series with the 10 mA source), but our voltmeter reads 0.00999999 V. So we estimate the resistance of our DUT to be about 0.999999 ohms. 

The error in the 2-wire case is about 2%, dominated by the series resistance of the interconnect wires. The error in the 4-wire case is only about 0.0001%, from the effect of the 1 megohm voltmeter in parallel with the load.