How to create variable power source circuit from a battery source?
Since a battery is like a voltage source with a resistor in series, to control the power you need to control the voltage.


Source: https://evertiq.com/design/33082


  so how does the design of the inductor affect the current?


The inductor switches on and transfers the current from the battery to "charge" the magnetic field of the inductor. When the switch is turned off (depending on the configuration of boost converter) the voltage on the inductor goes higher than what you could normally get from Vcc itself. 

Usually boost converters control voltage as shown in the feedback newtork above, usually in the form of a voltage divider. An error amlfiier ensures that the voltage of the feedback network is maintained, so if the attenuation of the resistor network was 1/12th and the error amplfiier was built to control to 1V, you would get 12V at the output of the diode (or secondary switch) in the boost converter. 

Voltage is usually what is controlled, and then the load determines the current, so if you have a voltage of 12V, and a 10Ω you'll get 1.2Amps of current. The boost converter doesn't care how much current, it's only trying to control the voltage. Some boost converters also have a current limit, meaning it won't allow more than X amount of amps. These boost converters also have a current resistor and amplifier to measure the current. If the current hits the limit then  the voltage drops on the output of the boost convert or it shuts off. 


  How does the total current to the load get determined? If for example,
  I have a load at 1ohm, and supply voltage (without boost) of 3.7v,
  then I get an output current of 3.7A. Now if I use the boost circuit,
  I can get an output of 10v, so is my output current 10A? 


Since it's probably voltage controlled you use V/R = I to determine the current. So 10V/1Ω is 10A. 


  Do I draw more current from the battery to generate the extra voltage
  since the battery is drained from both the 1ohm load and the inductor?


Energy must be conserved in circuits, if you had a perfect boost controller (100% efficiency) and you drew 10A at 10V then that would be 100W, the boost controller would need to get 100W from the battery, so if the battery were 3.7V you would get 100W/3.7V=27A of current (if it could actually source that much). 

Now suppose the converter is 80% efficient, to source 100W at the output you would need 100W/80% = 125W (you divide because your working backwards. 125W/3.7V=33.7A to run your boost converter.

There is one more caveat, the inductor saturates and has a limit to how fast it can switch, the boost converter is limited by the inductor on how much current (and sometimes the switches) can be transferred through the converter. 


  Will changing the switching frequency actually modify the output
  power? As far as I know, this is how commercial circuits basically
  work (for ecigarettes/vape mods, etc), is there some other way?


Yes an no, the switching frequency is determined by the controller which is attempting to regulate a voltage, the way to build boost converters is to carefully calculate the on and off time and control loop. An easy way is to buy an off the shelf converter. To build good DC to DC converters you need a knowledge of circuits and control theory.