How does the ionic strength formula work? (simple version)
Firstly; why is there a factor of a half out the front?


To avoid double-counting. For a salt such as $\ce{NaCl}$ comprising a monovalent cation and a monovalent anion, the ionic strength is defined such that it should be equal to the molar concentration of the salt, as opposed to the sum of both the anion and cation concentrations.  Therefore, since such a salt has two constituents, each constituent contributes half of the total ionic strength.


  Secondly; why do we square the charge number of the ion?


For this there's not much choice but to turn to the Poisson-Boltzman equation, which provides a differential expression for the electric field inside the electric double-layer. (Sorry, 5-year-old version of Jekowl!) Below is the version I have in my graduate transport book, Eq. (15.4-6):

$$
\nabla^2\Phi = -{F\over \epsilon} \sum_i{z_iC_{i\infty}\exp{\left(-{z_iF\over RT}\Phi\right)}}
$$


$\Phi$ is the electric potential, $\mathrm{J} \over \mathrm{C}$ or $\mathrm{V}$
$F$ is Faraday's constant, $96,485\ {C \over \mathrm{mol}\ q}$
$\epsilon$ is the electrical permittivity, $\mathrm{C}\over \mathrm{V\ m}$
$z_i$ is the valence of ion $i$, $\mathrm{mol}\ q \over \mathrm{mol\ ion}$
$C_{i\infty}$ is the concentration of ion $i$ far from the surface of interest, $\mathrm{mol} \over \mathrm{m^3}$
$R$ is the ideal gas constant, $8.314\ {\mathrm{J} \over \mathrm{mol\ K}}$
$T$ is temperature, $\mathrm{K}$
$q$ is the elementary charge


If the magnitude of the electric field is small such that $\left|z_iF\Phi/RT\right|\ll 1$, the exponential can be expanded as a Taylor series:

$$
\nabla^2\Phi = -{F\over \epsilon} \sum_i{z_iC_{i\infty}\left(1-{z_iF\over RT}\Phi+\dots\right)}
$$

If you then pull everything inside the parentheses, ignoring signs the second term becomes:

$$
C_{i\infty}z_i^2{F^2\over \epsilon RT}\Phi
$$

The second-order dependence of $z_i$ is readily visible in the above.