Let assume that we have two possible event $a_1$ and $a_2$ that may happen with probability $p(a_1)$ and $p(a_2)$ with $p(a_1)+p(a_2)=1$. Now let us consider a detector which is affected by some noise effects. Hence, we have a certain probability of correctly discriminate $a_1$ from $a_2$ and a certain probability of incorrectly discriminate the two cases: if we denote by $p^d(a_j \to a_i)$ the probability of the detector of observing $a_i$ given that the event was $a_j$, then the probability of error is \begin{equation} p_e^d = p(a_1)*p^d(a_1 \to a_2) + p(a_2)*p^d(a_2 \to a_1) \end{equation} i.e. the probability of having $a_1$ but the detector see it as $a_2$ plus the converse case.
So far the derivation should be correct, but if there are some hidden assumption or some mistakes, please tell me.
The next step is consider the following scenario. Let us assume that we have two detector affected by two independent noise and that can make two independent and simultaneous measure. Then we will have two $p^{d_1}_e$ and $p^{d_2}_e$ probabilities of error if we observe individually the two measurement. But what happens if I want to evaluate the probability of error of the joint observation of the two schemes $d_1$ and $d_2$? Is the probability of error (J is for joint) $p_e^J = p_e^{d_1}*p_e^{d_2}$? Or it is equal to the following one: \begin{gather} p_e^J = (p(a_1)*p^{d_1}(a_1 \to a_2)p^{d_2}(a_1 \to a_2)+p(a_2)*p^{d_1}(a_2 \to a_1)p^{d_2}(a_2 \to a_1)) +\\ + (p(a_1)*p^{d_1}(a_1 \to a_1)p^{d_2}(a_1 \to a_2)+p(a_2)*p^{d_1}(a_2 \to a_2)p^{d_2}(a_2 \to a_1))+\\ +(p(a_1)*p^{d_1}(a_1 \to a_2)p^{d_2}(a_1 \to a_1)+p(a_2)*p^{d_1}(a_2 \to a_1)p^{d_2}(a_2 \to a_2)) \end{gather} where the first line is the probability of having the word $a_i$ and erroneously see it as $a_2$ in both the detector; the second line is the one in which the $d_1$ detector correctly see the event while the $d_2$ does not; the third line is the one in which the $d_2$ detector correctly see the event while the $d_1$ does not.
What is the correct expression?
N.B.: please notice that $p(a_i \to a_j) + p(a_i \to a_i) = 1$.
The first step is correct without further assumptions.
For the second step, note that the joint error is not $p^{d_1}_e*p^{d_2}_e$: this is the probability of both detectors being wrong. Instead, you want the probability that at least one is wrong.
A simple way to calculate this, is to notice that the joint measurement is correct if and only if both measurements are correct. This has probability $(1-p^{d_1}_e)(1-p^{d_2}_e)$. So the probability to make a mistake is the probability that this does not happen: $$p^J_e=1-(1-p^{d_1}_e)(1-p^{d_2}_e).$$
Note that in your second derivation, you make a mistake: you need to distinguish between the two noises, since each detector has its own noise. That is, you need to have an $a_1^1$ and $a_1^2$ instead of only $a_1$.
PS: I assume that we are in the setting that both measurements measure a different noise effect and that the noises and measurements are independent. If you measure the same effect twice with different measurements, they should no longer be independent and your derivation needs to take the dependence into account.