I think there's some problem with Latex in your comment, I can't see it correctly. But I realized that you can indeed understand as conditional independence what I did in the numerator. I just did several steps in between, but you can apply conditional independence in the numerator to get to the last result faster. I didn't call it conditional independence because such a thing is not a fundamental law of probability, it is something that occurs naturally once you apply the product and independence rules.

Fixed. See edit.

Conditional independence is $P(d_1,d_2|x)=P(d_1|x) P(d_2|x)$. I don't do that in the numerator, I just use the product rule for joint probability. The relations I use are: $P(d_1, d_2 | x) = P(d_2, d_1 | x)= P(d_2|d_1,x) P(d_1|x)$ and $P(d_1|x)P(x) = P(x|d_1)P(d_1)$. In the denominator we have $P(d_1,d_2)=P(d_1|d_2)P(d_2)$ and then I use independence i.e., $P(d_1,d_2)=P(d_1)P(d_2)$.

I'm not sure I get your question. My final expression applies if $d_2$ and $d_1$ are independent of each other. If they are not independent of each other, you need my penultimate expression, but that's not all since in that case, you need to know how $d_1$ conditions $d_2$. Maybe you can reword your question and I can help.