PCB Current Draw vs Temperature Rise Interpretation and Question
Does this mean that the same amount of cooper in pcb (same trace width) actually can handle and conduct higher 80A current but the pcb temperature will increase 30°C more as a cost. Is this correct.


Yes. 


A trace will have a certain resistance and power dissipated will be given \$ P = I^2 R \$.
Increasing the current by \$ \frac {80}{50} \$ will result in a heating power increase of \$ \frac {80^2}{50^2} = 2.56 \$ or 256%.
It seems reasonable to assume that the final temperature rise would be 256% higher than that at 50 A but the vendor claims 300%.
Equilibrium temperature will occur at the point where heat lost to the surroundings = electrical power into the resistance. Generally the higher the temperature difference the faster the outflow of heat. This again suggests that the 300% figure is higher than we would expect.
The copper resistance has a positive temperature coefficient of +0.393%/°C. You can work out how much extra heat this will generate.
Don't forget that with the resistance of the track comes a voltage drop proportional to the current.



  If so do we need continuous amount of current draw to reach that temp rise. (If we draw 80A for short period of time like a peak does it still increase 30°C.)


No. You could make an approximation based on the duty cycle for a &lt; 1 s pulse (for example) but as the on-time gets long enough to be thermally significant you would want to assume the higher figure - a 10 s pulse for example.

See also: Standard PCB trace widths?.