A language is decidable iff it is Turing-recognizable and co-Turing-recognizable (WHY?)
Since both $A$ and $\overline A$ are recognizable, you know that for any input $x$, either

$x \in A$, which means that $T_A$ will halt on input $x$, or
$x \notin A$, which means that $x \in \overline A$ which means that $T_{\overline A}$ will halt on input $x$.

The &quot;trick&quot; here is to run $T_A$ and $T_{\overline A}$ &quot;simultaneously&quot;, which means that if any of $T_A$ or $T_{\overline A}$ halts after $k$ steps, then your new machine will halt after $2k$ steps.

If you don't like &quot;simultaneously&quot;, you can for each integer $i$ from 0 and up, run $T_A(x)$ for $i$ steps, then run $T_{\overline A}(x)$ for $i$ steps.
The latter is simpler to &quot;program&quot; on a single-taped Turing machine, but the former is quite okay to implement on a two-taped Turing machine.