Predict melting point of water CaCl2 solution
I think there are three issues here. The first is that your purple line of calculating the freezing point doesn't appear to be correct. Let's arbitrarily select $b=1 ~ \mathrm{mol/kg}$ to compare. $i=3$ and $K_\mathrm{f}=1.86 ~ \mathrm{^{o}C/mol}$. The molecular weight of $\ce{CaCl2}$ is $111 ~ \mathrm{g/mol}$. Then, our mass concentration is $b·\mathrm{MW}/(b·\mathrm{MW}+1 ~ \mathrm{kg})=0.1 ~ \mathrm{kg}_\mathrm{salt}/\mathrm{kg}_\mathrm{total}$. The freezing point is $b·i·K_\mathrm{f}=1·3·1.86=5.58 ~ \mathrm{^{o}C}$. This is much closer to the red/black line than the purple one.

The second problem is that that relation is only valid at low concentrations. As the concentration of salt goes up, it has an increasing effect on the activity of the water. As we can see from the phase diagram, once you get to a 6:1 mol ratio of $\ce{Ca^{2+}}$ to $\ce{H2O}$, the interaction is extremely strong. I'm just speculating here, but I would guess six water molecules are forming a weak octahedral coordination complex with the $\ce{Ca^{2+}}$, with the electronegative oxygen ends oriented toward the ion. This effect becomes so strong, that the hydrate becomes the relevant element for consideration above that concentration.

Third, you're ignoring the existence of hydrates. If you find the melting points and the cryoscopic constants of the hydrates, you can actually calculate the right side of the plot the same way.