What would happen to the voltage or current flowing through the motor in an h-bridge
Let's give some simple calculations based on some fictional values.


Motor resistance, Rm, = 1&Omega;.
Supply voltage, Vs, = 12 V.
With the applied load the motor runs up to the speed where the back EMF, Vb, is 11 V.


From these figures we can calculate:


Starting current is \$ \frac {V_s}{R} = \frac {12}{1} = 12 \ \text A \$.
Running current is \$ \frac {V_s - V_b}{R} = \frac {12 - 11}{1} = 1 \ \text A \$.


Now reverse the motor by reversing Vs.


Running current is \$ \frac {-V_s - V_b}{R} = \frac {-12 - 11}{1} = -23 \ \text A \$.



  ... if I suddenly change the rotation from clockwise to anticlockwise or vice verse in an H-bridge configuration, would the current flowing through double?


Worse. It would be almost double the starting current but in the opposite direction.


  Would the voltage double across the motor? 


No. It would just switch polarity. This assumes that the voltage supply is "stiff" and can absorb any current injected into it by the decelerating motor. Most practical systems will require some way of burning off the regenerated energy by switching in a brake resistor when the bus voltage gets too high. This will dissipate the energy as heat.