What is the relationship between the number of transition rules for an NDTM and the resulting number of computational branches? How can an NDTM have a growing number of branches as you feed larger and larger inputs with only finite number of transition rules specified--ie what is the relationship between the number of branches and the number of transitions?
An NDTM has a finite number of states, a finite number of input symbols and a finite number of transitions--but presumably the number of branches used grows as you feed larger and larger strings to it--how, mechanically, does this work?
if the number of branches keeps growing as word size grows does this not imply an infinte number of transition rules need be spelled out in the definition of NDTM (since there are an infinte number of word sizes) or do # of transition rules and # of branches not have a one-to-one relationship like this?
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.