What is the relationship between the number of transition rules for an NDTM and the resulting number of computational branches?
we can see how the computation tree grows with finite alphabet and rules even when using a DFA , consider the following :

We begin at 1 , then when reading a , we arrive at 1 and 2 , reading another a , 1 arrives at 1 and 2 , 2 arrives at 1 and 2 , so we are at 1,2,1,2 and so on , here is a part of the computation tree :

Hopefully you can see how the computation tree grows exponentially with every symbol , the arrows to the left represent the string read up to this level
Of course the same can be shown for a TM , you can imagine this DFA as a TM , which reads a symbol , writes nothing then goes to right , and you will yield same result
As for a relation , we know that every t(n) time nondeterministic single-tape Turing machine has an equivalent 2O(t(n)) time deterministic single-tape Turing machine.