In the first scenario, we have no reason to believe that the three outcomes stated are all equally likely. Why should two heads be as common as one of each? We must be careful not to assume that just because we can list three outcomes that they’re three equally likely outcomes.

In the second scenario, we can reason that hh must be just as likely as ht, because once coin $1$ has landed heads up, coin $2$ is just as likely to land heads up as tails up. Similarly, ht must be just as likely as tt because given that coin $2$ lands tails up, coin $1$ is just as likely to land heads up as tails up. In this way we can show that all four outcomes are equally likely.

Scenario one is actually no different from scenario two. It doesn’t matter if we’ve actually marked the coins as different; they are different coins regardless. So getting ‘one of each’ must be twice as likely as getting ‘both heads’, because it’s actually two outcomes: ht and th (even if there’s no difference between the two coins that we can see and we can’t actually tell which of these two outcomes has happened).