As is the case with ordinary multiplication, tensor product distributes over addition, so we can pull $|0\rangle$ on the first qubit out in front
$$
\begin{align}
|\Psi⟩ &= \frac{1}{\sqrt{2}}|\color{red}{0}0\rangle+\frac{i}{\sqrt{2}}|\color{red}{0}1\rangle \\
&= \frac{1}{\sqrt{2}}\color{red}{|0\rangle}\otimes|0\rangle+\frac{i}{\sqrt{2}}\color{red}{|0\rangle}\otimes|1\rangle \\
&= \color{red}{|0\rangle}\otimes\left(\frac{1}{\sqrt{2}}|0\rangle+\frac{i}{\sqrt{2}}|1\rangle\right) \\
&= \color{red}{|0\rangle}\left(\frac{1}{\sqrt{2}}|0\rangle+\frac{i}{\sqrt{2}}|1\rangle\right)
\end{align}
$$
and what remains in parenthesis is the state of the second qubit.
Note that people generally tend to make tensor product signs $\otimes$ implicit. I marked them explicitly to highlight the distributive law familiar from ordinary multiplication.