I have to calculate the $IDFT$ for a signal $y_2[n]$:
\begin{align*} y_2[n] = DFT^{-1} \Big\{ \Im m \{ \tilde{X}[k] \} \Big\} \end{align*}
and I am allowed to use some formulas from a collection like:
\begin{align*} \frac{1}{2}(x[n] - x^*[(-n) \text{mod} N]) = DFT^{-1} \Big\{ j\Im m\{ X[k]\} \Big\} \end{align*}
What does this $*$-operator mean and what's it doing here?
The $*$ means complex conjugation here. If $x[n]$ is real-valued you of course have $x[n]=x^*[n]$. The formula you quoted (wrongly, see below) comes from the fact that
$$\Im\{X[k]\}=\frac{1}{2j}\left(X[k]-X^*[k]\right)\tag{1}$$
and
$$DFT^{-1}\{X^*[k]\}=x^*[-n]\tag{2}$$
(I'm leaving out the modulo-$n$ operation here; it's always understood implicitly). Combining (1) and (2) gives
$$DFT^{-1}\{\Im\{X[k]\}\}=\frac{1}{2j}\left(x[n]-x^*[-n]\right)\tag{3}$$
If you compare (3) to the formula in your question, you'll note that there are a few typos in your formula.