Why are switching losses proportional to switching frequency?
The key to your question is 'soft switching'. So every 'edge' in the output there is a period of time when the output device is partially on- instantaneous power loss will, of course, be the current multiplied by the voltage across the device if we ignore any parasitic reactances. 

You can read more about the actual calculation of MOSFET losses here, but the key thing is that the loss is a little package of energy that has to be dissipated that occurs once per edge. The more edges in a second (higher frequency), the more total energy per second (which is power). 

From the above AN: