How does adding two Resistors in Series contribute to the sum of its resistances
how does the first resistor's resistance influences the second
  resistor's resistance.


It doesn't.  However, the resistance of the first resistor influences the voltage across the second resistor.

Clearly, the resistors in the diagram are series connected thus the current through each resistor is identical.

$$I_1 = I_2 = I$$

By Ohm's law, we have

$$V_1 = I_1 \cdot R_1 = I \cdot R_1$$

$$V_2 = I_2 \cdot R_2 = I \cdot R_2$$

By KVL, we have

$$6V = V_1 + V_2 = I \cdot R_1 + I \cdot R_2 = I \cdot (R_1 + R_2)$$

Thus

$$I = \frac{6V}{R_1 + R_2}$$

In other words, the current \$I\$ depends on the sum of the resistances (series connected resistances add).

The voltage across the second resistor can now be written as

$$V_2 = I \cdot R_2 = \frac{6V}{R_1 + R_2} \cdot R_2 = 6V \frac{R_2}{R_1 + R_2} $$

and so, as first stated, the resistance of the first resistor influences the voltage across the second resistor.

Similarly

$$V_1 =  I \cdot R_1 = \frac{6V}{R_1 + R_2} \cdot R_1 = 6V \frac{R_1}{R_1 + R_2}$$