Seen here, at the bottom of page 5, $\operatorname{Flatten}(\vec{a})$ is defined as:
$\operatorname{Flatten}(\vec{a})=\operatorname{BitDecomp}(\operatorname{BitDecomp}^{-1}(\vec{a}))$
For an n-dimensional vector $\vec{a} = (a_{1,0},\ldots,a_{1,l}, \ldots, a_{k,0},\ldots,a_{k,l-1})$. Where $a_{i,j}$ is the $j$-th bit in $a_i$'s bit representation.
But if we're computing the function through its own inverse, would this not just result in the original vector a?
When all the entries of $a$ are binaries, then yes, $Flatten(a) = a$, because $BitDecomp^{-1}(a)$ will give a $k$-dimensional vector whose entries are decomposed back to $\ell$ consecutive entries of $a$ (by $BitDecomp$).
However, when $a$ has non-binaries entries, this is not the case.
For instance, for $\ell = 3$ and $k = 2$, we would have $N = 6$. Then, consider $a = (0, 3, 0, 0, 0, 1)$.
Then, $BitDecomp^{-1}(a) = (3\cdot 2, 1\cdot 4) = (6, 4)$, and $BitDecomp(6, 4) = (0, 1, 1, 0, 0, 1)$.
Notice that for the last block of $a$ (last $\ell$ entries), $Flatten$ worked as the identity, because this block was binary.
