Equivalent resistances with the superposition theorem
If you simply follow along closely in the example in your book, it will make more sense.  In your picture they have just done the first half of solving by Superposition by short-circuiting V2.  

You understand why they use Ohm's Laws, but to understand what resistance values to use, think about it this way:

For i'1

With V2 shorted, r2 and r3 are now just two resistors in parallel, since both their end terminals are connected.  Do you see that?  You should now easily recognize the form of the equation for two resistors in parallel:
Rparallel = Ra * Rb / (Ra + Rb)
and now substituting r2 and r3 for Ra and Rb it becomes more clear.

Then you see that the parallel combination we solved above is in series with r1.  This total resistance has V1 dropped over it, hence the equation for i'1 using Ohm's law.

For i'2

This one may be a little harder to see, but basically you are forming a current divider with the resistors.  I challenge you to go over that section in your textbook again, or read through this carefully.

Keep at it and you will start to get it.