De morgan's law in formal language
Let $\Sigma$ be the alphabet on which $A_{TM}$ is defined over, and assume w.l.o.g that:

$A_{TM}\cup \overline{A_{TM}} = \Sigma^*$, and
$0, 1 \notin\Sigma.$

The set $J$ can be written as:
$$
J = \{0\}\cdot A_{TM} \ \cup \{1\}\cdot \overline{A_{TM}} 
$$
where $J$ is defined over the alphabet $\Sigma \cup \{ 0, 1\}$.
Now, using De morgan's law, it holds that:
$$
\overline{J} = \overline{\{0\}\cdot A_{TM}} \ \ \cap \overline{\{1\}\cdot \overline{A_{TM}} } \\
$$
$$
=\left(\{w: \text{ $w \neq \sigma\cdot u$, $\sigma\in\{0, 1\}$ and $u\in \Sigma^*$} \} \cup \{0\}\cdot \overline{A_{TM}} \ \cup \{1\}\cdot \Sigma^*\right) \ \cap \\
\left(\{w: \text{ $w \neq \sigma\cdot u$, $\sigma\in\{0, 1\}$ and $u\in \Sigma^*$} \} \cup \{1\}\cdot A_{TM} \ \cup \{0\}\cdot \Sigma^*\right) \\
$$
$$ = \{w: \text{ $w \neq \sigma\cdot u$, $\sigma\in\{0, 1\}$ and $u\in \Sigma^*$} \} \cup \left( \left[\{0\}\cdot \overline{A_{TM}} \ \cup \{1\}\cdot \Sigma^* \right] \cap \left[ \{1\}\cdot A_{TM} \ \cup \{0\}\cdot \Sigma^*\right] \right)
$$
$$
= \{w: \text{ $w \neq \sigma\cdot u$, $\sigma\in\{0, 1\}$ and $u\in \Sigma^*$} \} \cup \left( \{ 0\}\cdot \overline{A_{TM}} \cup \{ 1\}\cdot A_{TM} \right)
$$
So, we have that $w \in \overline{J}$ iff $w$ does not encode a bit followed by a word of the form $\langle M, w \rangle$, or $w = 1x$ for some $x \in A_{TM}$ or $w = 0y$ for some $y \in \overline{A_{TM}}$.
Finally, note that they're implicitly assuming w.l.o.g that every nonempty word in $\{0, 1\}\cup \Sigma^*$ is of the form &quot;$\text{bit} \cdot \langle M, w\rangle$&quot;. This assumption is okay, as checking whether a word respects some encoding can be easily done by a TM.