Yes, you can simulate several TMs on separate tapes and mark a separate tape with a 1 whenever one of your simulations halts. For TMs that are well behaved (like deciders), this approach will work and your approach will answer Yes correctly.
The immediate problem is that some TMs don't behave nicely—they will run forever without halting. No matter how many separate tapes you have, or how cleverly you divide up the simulation work, you will eventually encounter a TM that loops forever and uses up all of your tape resources. This means that your machine will not be able to say "No" because it is stuck waiting to find out.
The deeper problem is that the question "Does machine $M$ halt on at least $n$ different inputs?" is undecidable. If you had a subroutine that could decide it, you could solve the halting problem. (Proof below.) It's not just that your counting approach is flawed—no algorithm exists that decides this problem.
Proof that the problem is undecidable:
Suppose you had a machine $f(M,n)$ which decides whether $M$ halts on at least $n$ different inputs. Here's an algorithm for solving the halting problem:
Given a machine $M$ and a word $w$, design a new machine $M^\prime$. The machine $M^\prime$ is a modified version of $M$. If you give $M^\prime$ the input $w$, it simulates $M(w)$ as usual and halts if $M$ halts. On any other input, $M^\prime$ loops forever.
You can now solve the halting problem by computing $f(M^\prime, 1)$, which will return YES if $M$ halts on input $w$, or NO if $M$ doesn't halt on input $w$. This is a contradiction, so the algorithm $f$ can't exist.
