Turing Machine 'marking' specific portion of encoding
A quick answer: if the tape alphabet is large enough ($\geq 18$ symbols), then $L = \{\langle M,w\rangle \mid $ M on w  visits at least two unique states different than the starting state $\}$ is undecidable; indeed for every $M$ with $n$ states there is an equivalent $M'$ with only 2 states such that  for all $x$ $M(x)$ halts if and only if $M'(x)$ halts. To prove that L is undecidable, you can use the following reduction from the halting problem: given $\langle M\rangle$ build the equivalent two states machine $M'$ (the starting state plus another one), and add an additional final and accepting third state that can be reached only if $M'$ halts when run on the empty tape. Clearly $\langle M',w\rangle  \in L$ (i.e. visit the second and the third accepting state) if and only if $M$ halts ($w$ is irrelevant), so if $L$ is decidable you can solve the halting problem which is a contradiction.

Reducing the number of symbols, doesn't help too much, indeed you can encounter Turing machines that behaves like the Collatz function (e.g enter the third state only when the sequence reaches 1) ... and it is an open problem if for all $n$ the Collatz function reaches 1.

For a proof that you can build an Universal Turing machine (using 2-tag simulation) with only 2-states and 18 symbols see:

Y. Rogozhin. Small universal Turing machines. Theoretical Computer Science,
168(2):215–240, Nov. 1996.

Finally, as an exercise, you should be able to easily prove that the following language is undecidable:

$L' = \{ \langle M,w,k \rangle \mid$ when run on w the Turing machine M visits at least $k$ different states $\}$