In Shor's algorithm, how is ${\rm QFT}_n|x\rangle$ split into its even and odd components?

We can think of the set $\mathbb{Z}_{2^n}$ as the set of all binary sequences of length $n$ and we can write it as

$$ \mathbb{Z}_{2^n} = S^n_0 \cup S^n_1\tag1 $$

where $S^n_b=\{(y',b) |y\in\mathbb{Z}_{2^{n-1}}\}$ is the set of binary sequences of length $n$ that end in $b$, for $b\in\{0,1\}$. Note that $S_0\cap S_1=\emptyset$. Moreover, if $y\in S^n_0$ then

$$y=(y', 0) = 2y'\tag2$$

for some $y'\in\mathbb{Z}^{2^{n-1}}$ and if $y\in S^n_1$ then

$$y=(y', 1) = 2y'+1\tag3$$

for some $y'\in\mathbb{Z}^{2^{n-1}}$.

Using the observations above we can clarify the transformation in question as

$$ \begin{align} QFT_n |x\rangle &= \frac{1}{\sqrt{2^n}} \sum_{y \in \mathbb{Z}_{2^n}} \omega^{xy}|y\rangle \\ &= \frac{1}{\sqrt{2^n}} \left[ \sum_{y \in S_0} \omega^{xy}|y\rangle + \sum_{y \in S_1} \omega^{xy}|y\rangle \right] \\ &= \frac{1}{\sqrt{2^n}} \left[ \sum_{y' \in \mathbb{Z}_{2^{n-1}}} \omega^{2xy'}|y'0\rangle + \sum_{y' \in \mathbb{Z}_{2^{n-1}}} \omega^{x(2y'+ 1)}|y'1\rangle \right] \end{align} $$

where in the second step we used $(1)$ to split the sum and in the last step we used $(2)$ and $(3)$ to change the variable from $y\in\mathbb{Z}_{2^n}$ to $y'\in\mathbb{Z}_{2^{n-1}}$.