Power is current times voltage.
An ideal switch dissipates no power. When on the current may be high but the voltage drop is zero. When off the voltage may be high but the current is zero.
Transistor switches are not ideal, but the can come close. When on they have a high current and a low voltage drop, when off a small leakage current across a high voltage. They also don't switch directly from one state to the other but instead go through a period where the (effective) conductivity goes through intermediate values.
When the transistor is at some intermediate conductivity between full-on and full-off, it has a nontrivial current across a nontrivial voltage. So (unlike the full-on and full-off state) it dissipates a bunch of power. You'd like to keep out of this intermediate state as much as possible - going through it as rapidly as possible without causing other problems - to avoid frying the transistor.
With a purely resistive load (or a steady-state value, where the inductive and capacitive components of the load don't matter) you can treat the power supply plus the load as a "voltage source with a series resistance" and the "how much power does the transistor dissipate" question as one form of the impedance matching problem: The transistor will dissipate the most power when it has the same resistance as the load. If the load dissipates one unit of power when the transistor is a dead short, the load and the transistor will both dissipate 1/4 unit of power when the transistor is at the conductivity state where it dissipates the most.
When considering switching and a non-resistive load you have two additional issues:
As others have pointed out - when turning on the transistor you have an additional issue: It doesn't turn on over its entire area at once. Small regions start conducting first, then the conducting regions expand (and more small ones form and expand) until the whole thing is conducting. When the first regions turn on the total current is still low so the voltage remains high. But the current in the conducting regions is also high. So you're dissipating a lot of power in those small areas. This can cause them to be damaged if you hold this state for too long. So you want to turn on quickly, to get enough of the transistor conducting to pull down the voltage and reduce the power level in the parts that are conducting before you fry the tiny startup regions.
When turning off a load with an inductive component to its impedance the current can't change instantly. Instead the rate of change of the current is proportional to the voltage trying to change it. The result is much like inertia in a moving mass: It tries to keep moving and the quicker you try to stop it the harder it will push on you. With an inductor this produces a rise in voltage, called the "inductive kick". The magnetic field in the inductor represents stored energy, which must go somewhere if the field is to be eliminated. If you don't do something else about it it will increase the voltage across the transistor as it turns off so that the total energy is dissipated in the transistor.
However additional circuitry (called a "snubber") can be added to dump this power somewhere other than the transistor. A free-wheeling diode can let the current circulate through the coil until it's dissipated in the coil's resistance. A resistor in series with this can slow the current more abruptly and dump most of the energy in the resistor (rather than putting a spike of heat into the coil) - but the resistor must not be too large or the inductive kick will be high enough to force current through the transistor as it turns off. You could also dump the power back into the power supply (though this takes a double-ended power supply or more circuitry).
Presuming you have an adequate snubber circuit you're back to the same situation as a resistive load: You still want to turn off the current through the transistor quickly, to stay out of the high-dissipation, halfway-on state as much as possible and avoid overheating the transistor.
So turn it on fast and turn it off fast. Then you can control a lot of power with a low-power-dissipation transistor.
