Imagine a glass of water inside a box which is completely evacuated. There is no gas in the box at all, only liquid water and empty vacuum. Every so often, an H$_2$O molecule near the surface of the water gets a random kick from its neighbors which is sufficient to free it from the liquid phase and propel it outward into the vacuum. Through this mechanism, the liquid water evaporates and becomes water vapor.
Initially, there is hardly any vapor at all so the reverse reaction (an H$_2$O molecule in the vapor phase condensing back into the liquid phase) is rare. As more molecules enter the vapor phase, the rate of re-condensation increases until eventually the condensation rate is equal to the evaporation rate. At this point, the liquid is in equilibrium with the vapor, and the pressure of the gaseous H$_2$O is called the vapor pressure of H$_2$O. Note that vapor pressure is a function of temperature; if the glass of water is hotter, the rate at which H$_2$O molecules are kicked out into the vapor phase goes up, which means that the corresponding vapor pressure required for equilibrium will also go up.
If the box was not initially evacuated but rather filled with an atmosphere of, say, pure helium gas, then precisely the same process would take place. Therefore, the vapor pressure refers specifically to the partial pressure of H$_2$O which will be in equilibrium above its liquid phase, and is not concerned with the pressure due to other gases.
Now consider a pot of water on your stove. For the sake of argument, let's say your kitchen is completely sealed off from the outside environment except a piston which is exposed to the atmosphere; the purpose of this elaborate setup is to make sure that your kitchen stays at precisely one atmosphere, but no gas may enter or leave. Assume also for the moment that the gases in the kitchen are perfectly mixed.
As you increase the temperature of the water, the rate of evaporation increases - as a result, the vapor pressure increases as well. The proportion of H$_2$O vapor in your kitchen increases and the piston is slowly pushed outward, such that the system is always in equilibrium. At 99$^\circ$ C, the gas in your kitchen is almost all water vapor, but still an equilibrium is maintained.
But then you cross the 100$^\circ$ C point, and the vapor pressure exceeds $1$ atm. Runaway evaporation now occurs, as the partial pressure of water in your kitchen can never match the vapor pressure (which is the pressure required to maintain the evaporation/condensation equilibrium). This is boiling.
Now, this situation is obviously artificial. In real life, your kitchen is not sealed off, and the gases in the room are not perfectly mixed. The region directly above the surface of the water has a very high concentration of H$_2$O vapor, and it is in this region that the evaporation/condensation equilibrium is maintained - until the boiling point is reached, that is.