Say you want to prove your problem A is NP-hard and you know already that problem B is NP-hard. The fact that you can transform every instance of A to an instance of B does not prove anything, because it could be that you only create instances in a subset of B that are easy.
To prove that A is NP-hard, you have to do the opposite. Show how you transform every instance of B into an instance of A, so that a solution of A implies a solution of B. Then if you had a solver for A, you could use it to solve B, which is impossible unless P=NP (since B is known to be NP-hard). In this case, you have to transform TSP to Pac-Man, not the other way around. Some technicalities aside (which are important, nevertheless), this is how hardness proofs go, sorry if it was a bit obscure.
To prove that A is NP-hard, you have to do the opposite. Show how you transform every instance of B into an instance of A, so that a solution of A implies a solution of B. Then if you had a solver for A, you could use it to solve B, which is impossible unless P=NP (since B is known to be NP-hard). In this case, you have to transform TSP to Pac-Man, not the other way around. Some technicalities aside (which are important, nevertheless), this is how hardness proofs go, sorry if it was a bit obscure.