I’m assuming that any hand with at least two $10$’s and at least two hearts counts.
In the case in which you choose the $\heartsuit 10$ you’re overcounting the sets of $4$ cards that include more than two hearts. Consider for instance the set $\{\heartsuit 10,\clubsuit 10,\heartsuit 3,\heartsuit 7\}$: you count it once as
pick the $\heartsuit 10$, pick the $\clubsuit 10$, pick the $\heartsuit 3$, pick another card
and once as
pick the $\heartsuit 10$, pick the $\clubsuit 10$, pick the $\heartsuit 7$, pick another card.
You’re also overcounting sets like $\{\heartsuit 10,\clubsuit 10,\heartsuit 3,\spadesuit 10\}$, once as
pick the $\heartsuit 10$, pick the $\clubsuit 10$, pick the $\heartsuit 3$, pick another card
and once as
pick the $\heartsuit 10$, pick the $\spadesuit 10$, pick the $\heartsuit 3$, pick another card.
There are $\binom{39}3$ sets that contain the $\heartsuit 10$. $\binom{30}3$ of them contain no other heart, and $\binom{37}3$ of them contain no other $10$, so a first approximation to the number of sets that include the $\heartsuit 10$ and at least one other heart and one other $10$ is $\binom{39}3-\binom{30}3-\binom{37}3$. However, this undercounts: each set that contains no other heart and no other $10$ has been subtracted off twice. There are $\binom{27}3$ such sets, so the number of sets that include the $\heartsuit 10$ and at least one other heart and one other $10$ is actually
$\binom{39}3-\binom{30}3-\binom{37}3+\binom{27}3=234\;.$
If you’re counting only those hands with exactly two $10$’s and two hearts, your error is in saying that there are $37$ possible choices for the fourth card: there are only $27$, since it can’t be one of the remaining eight hearts or two $10$’s. In this case all you need do to get the right answer is change $37$ to $27$.
In both versions your count of the cases in which the $\heartsuit 10$ is not chosen is fine.