When two spheres of equal charge make contact, why does the larger sphere gain more charge?

The naive reasoning which leads to the conclusion that charges $Q_1$ and $Q_2$ of two touching conducting spheres with radii $R_1$ and $R_2$ are related by the relation $Q_1 = Q_2\frac{R_1}{R_2}$ is wrong. This formula holds only when the distance between the spheres $L$ is large compared to $R1$ and $R_2$, $L\gg R_1,R_2$, and the spheres are connected by a long wire.

The complete rigorous solution to this question can be found by using some formulas given in the book "Problems in Electrodynamics" by V.V. Batygin, I.N. Toptygin http://www.amazon.com/Problems-Electrodynamics-English-Russian-Edition/dp/0120821605 , for example see problems 211 and 67. In the notation used in that book, the case of touching spheres corresponds to taking the limit $a\to 0$ in the bispherical coordinates $(\xi,\eta,\alpha)$: $$ x=\frac{a\sin\eta\cos\alpha}{\cosh\xi-\cos\eta},\quad y=\frac{a\sin\eta\sin\alpha}{\cosh\xi-\cos\eta},\quad z=\frac{a\sinh\eta}{\cosh\xi-\cos\eta} $$ and solving the corresponding boundary problem. The surfaces of two almost touching spheres will have coordinates $\xi_1=\frac{a}{R_1}\to 0$,$\ \xi_2=-\frac{a}{R_2}\to 0$. Without going into all the details the final answer is $$ \frac{Q_1}{Q_2}= \frac{R_1}{R_2}\frac{1+2\int\limits_0^\infty\displaystyle\frac{e^{-t}\cosh[(1+k)t]-e^{-kt}}{\sinh[(1+k)t]}e^{-t}dt}{1+2\int\limits_0^\infty\displaystyle\frac{e^{-t}\cosh[(1+1/k)t]-e^{-t/k}}{\sinh[(1+1/k)t]}e^{-t}dt},\quad \text{where}\ \ k=\frac{R_1}{R_2}. $$

But if one is interested only in qualitative answer to the question, then as it was explained in the comment by D.W. above, it is energetically more favorable for the sphere with the largest radius to absorb more charge. It is the most obvious in the case when one of the spheres has much larger radius than the other $R_1\gg R_2$.

Consider an extreme case, where the two spheres both have just a little charge, and one of them is far larger than the other. Say one is molecule-sized and the other is Earth-sized, and they each just have two excess electrons. The two electrons on the small ball are quite close, and would prefer to be farther away from each other. So, when the large ball offers them some new space to lease, one of them gladly takes it, and the two electrons already on the large ball hardly mind at all.

In slightly more precise terms: the charges on the small ball, because they're closer together, repel each other more strongly than the net charge on the large ball repels them, so some of them jump over.