The clear stuff coming out of your kettle spout is 100% vaporized water. (I'll sidestep pedantic discussions of the meanings of "steam" and "water vapor".) It quickly entrains the cooler air around the kettle and mixes with it, forming a mixing cloud, which you can see and most folks refer to as "steam".
Now let's dilute that ejected 100% water vapor with some dry air at 20 C. The mixed result is cooler than 100 C and has a lower vapor pressure than 1 bar (because not all the stuff is water vapor any more). Once you've accounted for the different heat capacities of water vapor and water, and the thermal expansion or contraction of the constituents, you get the blue curve above. You read the curve like so: if I mix some amount of 20 C dry air into some 100 C water vapor, so that the result is 60 C, then it's partial pressure of water vapor is 0.29 bar. At least, that would be the partial pressure if all the water was in vapor form.
The really neat thing here is that the blue curve is ABOVE the red curve all the way down to 26 degrees C, which happens to correspond to 95.8% dry air and 4.2% water vapor. At any mixing ratio between this and 100% water vapor, there will be more water in the mixing cloud than can be in vapor phase, and some will be in the form of very tiny (micron) water droplets. However, once the mixing cloud is sufficiently diluted with dry air, there will be less water vapor in it than the maximum that can be in vapor phase, and all those tiny droplets with their huge surface air to volume ratio will immediately evaporate.
What this means is that the mixing cloud grows to 23 times bigger than the pure water vapor expelled from the kettle before it disappears.