The second approach is over-counting. To illustrate why, denote the defective balls by $D_1,D_2,D_3,D_4$ and the non-defective by $N_1,...,N_8$.
If we begin by choosing two of the defective balls, say $D_1$ and $D_2$, then yes, there are ten other balls total. However, if we pick another of the defective balls, say $D_3$, then we end up with the same combination as we'd have if we had started with $D_1$ and $D_3$ and then chosen $D_2$--or the same combination that we'd have if we had started with $D_2$ and $D_3$ and then chosen $D_1$.
Generalizing the subscripts above, there are $4\cdot 2=8$ extra combinations that we're counting twice. So, subtracting 8 from ${4 \choose 2}{10 \choose 1}$ we get 52, which is the same as ${4 \choose 2}{8\choose 1} + {4 \choose 3}$. :)
This is an excellent question that illustrates how even basic probability problems can be counter-intuitive.