power efficiency's definition
The short answer: both are actually the same representation.

Let's start by a few definition:

Input: what goes in the system

Output: what goes out of the system or more usefully: what power is actually used by the intended purpose (movement, lighting, data transmission...doesn't matter).

If a system was perfect and without loss, the input power would be equal to the output power. However, no system is perfect. Therefore a certain percentage of the input energy is lost in the process. Most often, that loss is in the form of undesired heat.

To go back to your equation. The input power (what goes in) is equal to the output power (what goes out) plus the imperfection of the system (what is lost).

\$P_{in} = P_{out} + P_{loss}\$

You merely change \$P_{in}\$ in your first equation (\$\frac{P_{out}}{P_{in}}\$) to get your second equation.

For the second part of the question, by doubling the voltage, you will most likely reduce the current by roughly half at the output level. Therefore \$P_{in} = V×I\$ and \$P_{out} = \not{2}V×\frac{I}{\not{2}} - P_{loss}\$.

Any imperfection in the system will simply add up to what is outputted by the system. Some circuits may increase the total power output by using external sources during your process...In that case, it is simply a matter of taking into account those power injection throughout your circuit. But a passive circuit or a circuit that doesn't inject power along the way will always have an output that is less than the input.