EQ_{TM} is not Turing recognizable, but we can reduce A_{TM} to it?
Note that $EQ_{TM}$ is the language of machines that recognize the same language (they are not necessarily equivalent).
You said

But we can also reduce $A_{TM}$ to $EQ_{TM}$ itself, that means: $EQ_{TM}$ and $NEQ_{TM}$ are not disjoint since we can reduce $A_{TM}$ to the two of them?

This is far from being true, if there are reductions from a language $A$ to two languages $B$ and $C$, this does not say anything about $B\cap C$ as the reductions need not be related. To be concrete, it is shown here that every non-trivial language is $\text{R}$-hard, so clearly, you cannot claim that every two non-trivial languages intersect.
You also said

And also, $EQ_{TM}$ is not Turing recognizable but it contains some instance that is turing recognizable since we can reduce $A_{TM}$ to it?

Clearly, $EQ_{TM}$ contains some instances $\langle M_1, M_2 \rangle$, where it is easy to check whether $L(M_1) = L(M_2)$ (for example, all instances where $M_2$ is identical to $M_1$ with some new unreachable states). Also, you're right, this indeed can be thought of a consequence of the existence of a reduction from $A_{TM}$ to $EQ_{TM}$. Formally, we have the following claim (I leave it to you as a good non-trivial exercise), and thus the image of the reduction (which is a subset of $EQ_{TM}$) is in $\text{RE}$.
Claim: Let $f: \Sigma^* \to \Sigma^*$ be a computable function and let $L\subseteq \Sigma^*$ be a language in $\text{RE}$. It holds that $f(L) = \{ f(x): x\in L \}\in \text{RE}$.
BTW, complement languages, by definition, have empty intersection.
So, regarding the question in the title, you can reduce $A_{TM}$ to $EQ_{TM}$ (here is a hint: there they asked about a reduction from the halting problem, but the hint is also good for your case), and being able to define such a reduction does not contradict anything.