Calculate the wavefunction step by step by applying each gate in sequence
$$
\begin{align}
|000\rangle&\xrightarrow{H\otimes H}\frac12\sum_{a,b=0}^1|a\rangle|b\rangle|0\rangle\tag1\\
&\xrightarrow{CU}\frac12\sum_{a,b=0}^1i^a|a\rangle|b\rangle|a\rangle\tag2\\
&\xrightarrow{CU}\frac12\sum_{a,b=0}^1i^{a+b}|a\rangle|b\rangle|a\oplus b\rangle\tag3\\
&\xrightarrow{CU}\frac12\sum_{a,b=0}^1i^{a+2b}|a\rangle|b\rangle|a\rangle\tag4\\
&\xrightarrow{QFT^\dagger}\frac14\sum_{a,b=0}^1i^{a+2b}\sum_{x,y=0}^1i^{-(a+2b)(x+2y)}|x\rangle|y\rangle|a\rangle\tag5\\
&=\frac14\sum_{x,y=0}^1\sum_{a,b=0}^1i^{-(a+2b)(x+2y-1)}|x\rangle|y\rangle|a\rangle\tag6\\
&=\frac14\sum_{a,x,y=0}^1i^{-a(x+2y-1)}\left(\sum_{b=0}^1(-1)^{b(x+2y-1)}\right)|x\rangle|y\rangle|a\rangle\tag7\\
&=\frac14\sum_{a,x,y=0}^1i^{-a(x+2y-1)}\left(1-(-1)^{x+2y}\right)|x\rangle|y\rangle|a\rangle\tag8\\
&=\frac12\sum_{a=0}^1\left[|1\rangle|0\rangle+(-1)^a|1\rangle|1\rangle\right]|a\rangle\tag9\\
&=\frac{1}{\sqrt2}(|1\rangle|0\rangle|+\rangle+|1\rangle|1\rangle|-\rangle)\tag{10}
\end{align}
$$
The first qubit is the least significant qubit of the first register, so the final expression implies 50% probability of measuring $.01$ (encoding eigenvalue $i$) and $.11$ (encoding eigenvalue $-i$), as expected, c.f. this answer. Note that if the first register is measured, then the second collapses to the eigenvector corresponding to the eigenvalue read out from the first.