Ignoring the distinction between unconditional and conditional probabilities, and what are the effects of conditioning on the probabilities, is where the conceptual issue lies here.
Assume the two coins are visibly numbered, $c_1$ with known heads probability $p_1$, and $c_2$ with known heads probability $p_2$. Neither coins are assumed, or not assumed, "fair".
Then if we are told that the throw returned "one heads and one tails", but we have not actually looked at the coins, then the best we can say is the same thing we would have said prior to the throw:
$$P[\{HT,TH\}] = p_1(1-p_2) + (1-p_1)p_2$$
Assume now that we observe the coins, and them being visibly marked, we know which one came up heads and which one came up tails. But now, if we are to take into account what we have just observed, we are talking for the probability of the said event conditional on the additional information due to observing the coins. So here, we are trying to calculate the probability, say,
$$P\Big[\{HT,TH\} \mid \{c_1=H, c_2=T\}\Big]$$
It should be evident that now we are calculating a different probability than before, and so in any case, we should not confuse the first situation with the second.
So let's focus on this second situation. Since we know which coin came up heads, we are no longer considering the event $\{HT,TH\}$, because it describes a situation where either of the coins came up heads, while we know which one actually came up heads. Therefore we conclude that
$$P\Big[\{HT,TH\} \mid \{c_1=H, c_2=T\}\Big] = P\Big[\{HT\}\Big] = p_1(1-p_2)$$
And this has nothing to do with what are the values $p_1$ and $p_2$, the "drop" in probability value does not happen due to their "varying probability", but because, by introducing additional information, we have restricted ourselves to consider a different event than without this information.
If this has been presented as some kind of conceptual "problem", "paradox" etc, it should not.
