This answer is going to be updated several times. Each update will reflect my using the Addendum, at the end of my answer to field the questions of Martin Smartly, one at a time.
Let $~A~$ denote the probability that none of the $~6~$ opponents has two hearts. Then, the desired computation is $~1 - A.$
For $~k \in \{0,1,2,3,4,5,6\},~$ let $~f(k)~$ denote the probability that exactly $~k~$ of the $~6~$ players have exactly 1 heart, and the other $~(6-k)~$ players have no hearts.
Then $~\displaystyle A = \sum_{k=0}^6 f(k),~$
so the entire problem reduces to computing $~f(k).~$
There are $~\displaystyle \binom{6}{k}~$ ways of choosing which $~k~$ players will be the ones with exactly 1 heart, so you have that $~f(k) = \displaystyle \binom{6}{k} \times g(k),~$ and the problem reduces to computing $~g(k).~$
Here, $~g(k)~$ represents the probability that the $~k~$ players to your immediate left are the ones with exactly 1 heart, and the other $~(6-k)~$ players are the ones with no hearts.
So, now the problem has been reduced to computing $~g(k).$
Per the posted question, it is assumed that there are $~45~$ cards remaining in the deck, of which $~10~$ are hearts.
To ease the analysis, I am going to express the probability as $~\dfrac{N}{D},~$ where
$$D = \binom{45}{2} \times \binom{43}{2} \times \cdots \times \binom{35}{2} = \frac{(45)!}{(33)! \times 2^6}. \tag1 $$
This means that I am regarding it as relevant which players got which cards. This implies that I have to compute the numerator $~N,~$ in a consistent manner. The sections below describe the computation of the numerator of $~g(k),~$ for the different values of $~k.~$
$\underline{\text{Computation for} ~k = 0}$
$$N = \frac{(35)!}{(23)! \times 2^6}. \tag2 $$
The computation in (2) above is identical to the one in (1) above, except that the $~10~$ hearts have been excluded.
$\underline{\text{Computation for} ~k > 0}$
The numerator will be expressed as
$$N_1 \times N_2~$$
where $~N_1~$ equals the number of ways that the $~(6-k)~$ players can be dealt no hearts, and $~N_2~$ equals the number of ways that the $~k~$ players can be dealt 1 heart each.
Piggybacking off the previous section,
$$N_1 = \frac{(35)!}{\{ ~35 - [2(6-k)] ~\}! \times 2^{6-k}}. \tag3 $$
In the previous section, you were dealing two non-hearts each to $~6~$ players. In (3) above, you are dealing two non-hearts each to $~(6-k)~$ players.
At this point, there are $~35 - [2(6-k)] = (23+2k)~$ non-hearts left, and $~10~$ hearts left.
Here, the number of ways of giving exactly 1 heart and 1 non-heart, to each of $~k~$ players is
$$N_2 = \left[ ~\binom{23+2k}{1} \times \binom{10}{1} ~\right] \\
\times \left[ ~\binom{22+2k}{1} \times \binom{9}{1} ~\right]
\\
\times \cdots \\
\times \left[ ~\binom{24+k}{1} \times \binom{11-k}{1} ~\right]
$$
$$= \frac{(23 + 2k)!}{(23 + k)!} \times \frac{(10)!}{(10-k)!}. \tag4 $$
$\underline{\text{Final Computations}}$
The probability of at least one person having two hearts is
$$1 - A.$$
$$A = \sum_{k=0}^6 f(k) = \sum_{k=0}^6 \left[ ~\binom{6}{k} \times g(k) ~\right].$$
$$g(k) = \frac{N}{D} ~: ~D = \frac{(45)!}{(33)! \times 2^6}.$$
For $~k = 0,~$ the corresponding numerator is
$$N = \frac{(35)!}{(23)! \times 2^6}. $$
For $~k > 0,~$ the corresponding numerator is expressed as
$$N_1 \times N_2,$$
where
$$N_1 = \frac{(35)!}{(23 + 2k)! \times 2^{6-k}} $$
and
$$N_2 = \frac{(23 + 2k)!}{(23 + k)!} \times \frac{(10)!}{(10-k)!}. $$
$\underline{\text{Addendum-1}}$
...I don't understand from the part where you stopped using variables and inserted concrete numbers instead. That is from the line D = (45 2) onwards.
Okay, one slow, careful step at a time. The first thing to do is clear up any confusion that you have regarding my computation of $~D.~$
For Combinatorics problems such as this, it is more convenient to express the probability as $~\displaystyle \frac{N}{D},~$ where $~D~$ represents the enumeration of the entire universe of equally likely possibilities, and $~N~$ represents the enumeration of the subset of satisfying possibilities.
When doing the enumeration, you are allowed to choose whatever convention is most convenient, as long as that convention is consistently applied in the computation of both $~N~$ and $~D.~$
For convenience, I arbitrarily decided that I was going to distinguish which player got which cards.
Per your posted question, at the time that the $~6~$ players each receive two cards, there are $~45~$ cards left in the deck, $~10~$ of which are hearts.
Identify the $~6~$ players as Player-1, Player-2, ..., Player-6.
Then, the number of possible pairs of cards that could be dealt to Player-1 is $~\displaystyle \binom{45}{2}.~$
Once Player-1 receives his pair of cards, there will then be $~43~$ cards left in the deck. At this point, the number of possible pairs of cards that could be dealt to Player-2 is $~\displaystyle \binom{43}{2}.~$
This implies that if you were only focusing on Player-1 and Player-2, then the number of ways that each of the two could be dealt a pair of cards is
$$\binom{45}{2} \times \binom{43}{2}. \tag5 $$
It is critical that you understand how the expression in (5) above was derived. No matter which pair of cards is dealt to Player-1, there will then be $~\displaystyle \binom{43}{2}~$ distinct pairs of cards that can be dealt to Player-2.
Assuming that you understand (5) above, you should now understand (1) above.
$\underline{\text{Addendum-2}}$
...Then I don't understand however, why we count N1 and N2 again, if in each g(k) we already know which players got the heart and which not....
First of all, I am going to assume that you understand the computation of the numerator for $~\displaystyle g(0),~$ which represents the number of ways that each of the $~6~$ players can be dealt $~2~$ cards each, where none of them got any hearts.
Remember that $~k~$ represents the number of players that got exactly 1 heart. This assumes that there were $~(6-k)~$ players who got no hearts.
For $~k \in \{1,2,3,4,5,6\},~$ in order to compute the numerator for $~f(k),~$ you must:
Determine how many ways there are of choosing the $~k~$ people (out of 6 people) who will be the ones who each got 1 heart.
The computation here is $~\displaystyle \binom{6}{k}.~$
Then, determine how many ways there are for $~(6-k)~$ people to each get no hearts.
Then, determine how many ways there for $~k~$ people to each get 1 heart.
For illustrative purposes, I am first going to illustrate the computation for the last two bullet points above in the specific case of $~k = 3.~$ Then, I will generalize the formula for any $~k \in \{1,2,3,4,5,6\}.$
Assuming that the $~3~$ people who will not get any hearts have been determined, the number of ways that these $~3~$ people can each get a pair of non-hearts is
$$N_1 = \binom{35}{2} \times \binom{33}{2} \times \binom{31}{2} = \frac{(35)!}{(29)! \times 2^3} \\
= \frac{(35)!}{\{35 - [2(6-3)]\}! \times 2^{6-3}}.$$
The reasoning behind the computation of $~N_1~$ above is very similar to the reasoning that was used in Addendum-1.
At this point, regardless of which $~6~$ non-hearts have been dealt to the $~3~$ players chosen to receive non-hearts, there will now be $~29~$ non-hearts and $~10~$ hearts left in the deck.
Then, the number of ways that the $~3~$ remaining people can each be dealt 1 heart + 1 non-heart is
$$N_2 = \left[ ~\binom{29}{1} \times \binom{10}{1} ~\right] \\
\times \left[ ~\binom{28}{1} \times \binom{9}{1} ~\right] \\
\times \left[ ~\binom{27}{1} \times \binom{8}{1} ~\right] \\
= \frac{(29)!}{(26)!} \times \frac{(10)!}{7!} \\
= \frac{[23 + (2\times 3)]!}{[23 + (1\times 3)]!} \times \frac{(10)!}{(10 - 3)!}.$$
The reasoning here is that there are $~3~$ people who are each going to get $~1~$ heart and $~1~$ non-heart. For the sake of convenience denote them as Player-1, Player-2, and Player-3.
The first two factors above represent the number of ways that Player-1 can receive his two cards. Once Player-1 receives his two cards, the deck will have changed to $~28~$ non-hearts, and $~9~$ hearts. So, the second factor represents the number of ways that Player-2 receives his two cards, after Player-1 has received his.
Then, with $~k = 3,~$ the product $~N_1 \times N_2~$ represents the product of :
You can see how this would generalize.
For general values of $~k \in \{1,2,3,4,5,6\},~$
$$N_1 = \frac{(35)!}{(23 + 2k)! \times 2^{6-k}} $$
and
$$N_2 = \frac{(23 + 2k)!}{(23 + k)!} \times \frac{(10)!}{(10-k)!}. $$
If you experiment with various other values of $~k \in \{1,2,3,4,5,6\},~$ following the approach that I took in this section, you will see that the formulas for $~N_1~$ and $~N_2~$ work.
