Confusion on electrode potentials
My question is that suppose we have two metals, A and B. A produces +1 ions. B produces +2 ions. Now even if the two have the same tendency to lose or gain electrons, the B plate would be 'more negative'.


That is not a problem because the tendency of the metal to lose or gain electrons has no direct bearing on the reactivity of the metal or the reduction potential. You are thinking of this on an "per atom of metal" basis, but the reduction potential is on an "per electron" basis. One way to see that is that the cell potential does not change when you double the coeffients in an equation:

$$\ce{Ag+ + 1/2Mg -&gt; Ag + 1/2 Mg^2+}$$

has the same cell potential as 

$$\ce{2Ag+ + Mg -&gt; 2Ag +  Mg^2+}.$$

(This is saying something like the potential of a D cell battery is the same as that of a AAA cell battery, even though the D cell battery is much bigger). On the other hand, the Gibbs energy of reaction would be twice as large for the second reaction compared to the first. One way to think about this is that the work the second reaction would do (if this is set up as a voltaic cell) is twice that of the first reaction because two electrons are transferred instead of one (across the same cell potential).


  How has the voltage helped us determine the difference in the tendencies of these two metals? 


What we really want to know is which of the two metals would be reduced and which would be oxidized. For that, we need to know the potential (or in your picture, the excess or deficiency of electrons in the respective electrodes). This will tell you the direction of the reaction, and the cell potential when connecting the two electrodes and the two electrolytes to allow the reaction to proceed.

You can measure the potential at next to no current, so the stoichiometry of the reaction does not matter at this point - you just have to consider the charge separation that already has happened at the electrodes before you connect them.