Why is dominating set in $W[2]$, but independent set in $W[1]$
For deciding the independent or dominating set problem, it is not sufficient to guess the set. One must also verify that the guessed set indeed is an independent or a dominating set.

In the case of independent set, once the machine has guessed the set and stored it in the first $k$ cells of the tape, it checks for all pairs of guessed vertices whether they are indeed not equal and do not form an edge. There are $k\choose 2$ such pairs, so the overall running time is $O(k^2)$. Thus the reduction from independend set to short NTM acceptance is a parameterized reduction. The number of steps in the verification phase only depends on the parameter.

In the case of dominating set, one needs to checks whether all vertices are dominated. While above, we only had $k\choose 2$ pairs, now we have to deal with $n$ vertices. Which does not only depend on the parameter. (The way the multitape NTM solves this is by spending one tape per vertex.)

Of course, the above is only intended to give some intuition about why dominating set is not in W[1]. It does not give a proof. That would be way out of scope...

(For the sake of the casual reader, let me clarify that the NTMs above are not machines that solve the independent or dominating set problem. There is a reduction involved that converts instances of these problems into machines. Thus, a single such machine solves only a single instance of the problem, not the general problem.)