Titration with Ca(OH)2
(1) Meaning of the textbook solution

$$\ce{2HCl + Ca(OH)2 = CaCl2 + 2H2O}$$

This reaction clearly tells us that 

$$\frac{moles_{acid}}{moles_{base}}=2$$

moles=molarity* volume(in litres)

Therefore, $$\frac{moles_{acid}}{moles_{base}}=\frac{M_{a}V_{a}}{M_{b}V_{b}}$$

(2) What is wrong with your interpretation?

You wrote $2M_\mathrm{a}V_\mathrm{a} = M_\mathrm{b}V_\mathrm{b}$.

$2M_\mathrm{a}V_\mathrm{a} = M_\mathrm{b}V_\mathrm{b}$ actually implies that more number of moles of base need to be taken. For neutralisation, for every one mole of base, we require two moles of acid since each calcium hydroxide will release 2 hydroxide ions (for which two H+ are required).