K-path sampling transfer function
Equation 4 follows directly from equation 3, since it can be rewritten as
$$
\frac{V_{out}(z^{1/K})}{V_{in}(z^{1/K})}=\frac{\sum_{n=1}^K V_{out,n}(z^{1/K})}{\sum_{n=1}^K V_{in,n}(z^{1/K})}.
$$
Then from equation 2 it follows that \$V_{out,n}(z^{1/K})=H(z)V_{in,n}(z^{1/K})\$, and since all paths are assumed to be equal (at least they are associated to the same transfer function) \$H(z)\$ can be taken out of the summation:
$$
\frac{V_{out}(z^{1/K})}{V_{in}(z^{1/K})}=\frac{H(z)\sum_{n=1}^K V_{in,n}(z^{1/K})}{\sum_{n=1}^K V_{in,n}(z^{1/K})}=\frac{H(z)V_{in}(z^{1/K})}{V_{in}(z^{1/K})}=H(z).
$$
In words, since by definition the individual paths' transfer functions can be expressed as \$H(z)=\frac{V_{out,n}(z^{1/K})}{V_{in,n}(z^{1/K})}\$ and at the output the circuit considers one such \$V_{out,n}(z^{1/K})\$ at the time (due to the non-overlapping clocks \$\phi_n\$), the transfer function that describes the overall circuit is the very same \$H(z)\$.