Why a TM with infinite states can decide the halting problem?
If you are allowed infinite states and transition functions, then even a DFA could solve the halting problem (and in fact, any language). In this case, the DFA/TM can basically have a state for each possible pair of TM + word. If a certain state represents a pair where the TM halts on the word it would be an accepting state.

Notice that I don't give a constructive algorithm of how to build this TM/DFA, but it could easily be seen that such a TM theoretically exists and therefore decides the halting problem.

EDIT - more detail regarding the last point: by definition, a language is regular if there exists some DFA accepting it, even if we don't know how to build that DFA. If we allow infinite states, than all languages will be regular. The infinite-state-DFA I described above may seem confusing because we have no idea how to construct it, and in particular, how to choose which states are accepting and which aren't. However, it definitely still exists! 

Another way to think about it is that the states of the DFA can be computed/built in "compile-time" and therefore we don't need to figure out whether a TM stops on a word in "run time". We determine what is accepting and what isn't before we even get any input, and when we do get an input, we will definitely halt on it because of the nature of DFAs. We don't need to simulate a TM on an input during the DFAs run!

Lastly, I want to emphasize that the confusion regarding DFAs which we have no algorithm to construct is valid even using the standard, finite-state model of DFAs. If we denote the halting problem by $H_{TM}$, then for all $n$, the language $H_n:=H_{TM}\cap \{0,1\}^n$ is a finite language and therefore regular! However, it can be proved that there is no general algorithm that takes $n$ as an input constructs a DFA for $H_n$, since otherwise we could decide the halting problem.