How can we directly add half cell potentials to measure the EMF of a galvanic cell?
You are writing $\Delta G_3=\Delta G_1+\Delta G_2$. But it should be $\Delta G_3=\Delta G_1+2\Delta G_2$ (I am considering  $\Delta G_2$ as free energy change for the conversion of $\ce{Ag+}$ to $\ce{Ag}$, which is considered 2 times in the final reaction) Because $\Delta G$ also depends on the number of moles of reactants and products, if you add two times the second reaction, the $\Delta G$ will also be used as twice its value. There you've made a mistake.

Thus the equation will be $$2E_3^0 = 2E_2^0 + 2E_1^0$$ So you will have $E_3 ^0 = E_1^0 + E_2^0$.