According to my textbook, in steam distillation the liquid boils when the sum of vapor pressure due to the organic liquid and that due to water becomes equal to the atmospheric pressure. Why? I did not understand.
Distillation of many organic oils and resins requires temperatures significantly higher than 100$^o$C. If you add water to the oil/resin (say, 50/50), the mixture will boil at 100$^o$C. Well, not quite 100$^o$C, because at the boiling temperature, the oil or resin will have a slight vapor pressure of its own, and the total vapor will therefore have some water molecules and some oil molecules, which together will exert a pressure of one atmosphere (plus a tiny tiny extra pressure that is needed to force this vapor out of the flask).
In practice, steam distillation is essentially using the water to provide nearly one atmosphere of pressure to carry the slightly volatile oil/resin out of the boiling flask. This is not normally done at equilibrium conditions, but at a rapid boil so you can get some product in a reasonable time. However, the theoreticians are correct that the water has a vapor pressure, and the oil/resin has a vapor pressure, and they theoretically add up to one atmosphere (plus whatever added pressure is needed to force the vapor out of the flask at the rate you are using).
Steam distillation is not the best description for a simple separation of water from an organic, nor for an azeotropic distillation, where water is added so a "compound" (azeotrope) of the water plus the organic can be carried over because its composition is more preferred than the original mixture.
Evaporation happens all the time, at a positive net rate as long as the vapor pressure of that component is larger than the atmospheric partial pressure of the same component.
Boiling is a specific case of evaporation, at the pressure and component specific boiling point, when the total pressure equals the atmospheric pressure. It is fairly trivial to show that at least one of the liquid components must have greater partial pressure than the atmosphere for this case to exist. At this state the temperature of the liquid cannot increase, until the situation is resolved. Either by system pressure rising or by elimination of the liquid.
It would be beneficial to read up on thermodynamics, phases and states, before proceeding with regular chem, it's only going to get worse...
Imagine for illustrative purposes there are two liquids with identical vapour pressure and boiling temperature.
Now consider two scenarios:
Liquids are miscible, in 1:1 molar ratio. Saturated vapor pressure of each is half of the vapour pressure of the pure liquid. But summed, it still makes vapour pressure of either of liquid, so the mixture boiling point hoping stays. For non equal boiling points, the vapor pressure would be the average of pure liquid pressure ( at ideal behaviour ).
Now, liquids are not miscible. Each liquids forms separate phase of the pure liquid and keeps the saturated vapor pressure of pure liquid. But there are 2 liquids, so there is twice as big total vapor pressure. Therefore the boiling occurs at temperature each of saturated pressures is just a half of the external pressure.
For highly non equal boiling points, total vapor pressure is still sum of pressures, with the mixture boiling point somewhat less than if the more volatile liquid.