Why should we multiply Copper loss with \$ (\frac{1}{2})^2 \$, not \$(\frac{1}{2})\$ when the transformer is in half loading?
Because \$P=I^2R\$ since Copper loss is resistive loss so it is proportional to current squared.

Half the loading means half the wattage which means half the current IF your voltage stays the same. After all, you changed the loading to reduce the wattage right? You didn't reduce the input voltage?

If you reduced the wattage such that the voltage was less but current is the same, then copper loss stays the same.