Why is the common magnesium ion Mg(II) and not Mg(I) when the second ionization energy is higher than the first ionization energy?
As mentioned by Zhe, we have to look at the entire process by which an ionic compound is formed, not just the energy for a single part.
For example, consider the formation reaction of $\ce{MgO}$
$$\ce{Mg(s) +1/2O2(g)-&gt;MgO}$$ To find the enthalpy of this process, we could follow a Born-Haber cycle: 
This demonstrates that while the ionization of $\ce{Mg}$ to $\ce{Mg^{2+}}$ might be unfavorable, the overall process to form the compound is not. So the ionization of $\ce{Mg}$ is driven by the favorability of the overall process.2
It is also worth noting why we do not instead form $\ce{Mg2O}$. If we were to construct a Born-Haber cycle for this compound, it would take less energy to form $\ce{2Mg^{+}(g) +O^{2-}(g)}$ than it would to form $\ce{Mg^{2+} (g) + O^{2-}(g)}$. However, the lattice energy of $\ce{Mg2O}$ would be sufficiently smaller than the lattice energy of $\ce{MgO}$ that $\ce{MgO}$ is still more favorable to form. Lattice energies are heavily dependent on the charge of the ions involved and having a $+2$ ion will lead to twice as large a lattice energy as an ion with a $+1$ charge all else being equal.
Even in this simple case, all else might not be equal. To make a more direct comparison between we can utilize the Born-Lande or Born-Mayer to obtain a calculated value of the lattice energy.