Understanding index transformation in derivation of Fourier transform for sampling rate reduction
It's not derived, it's just chosen in a smart way such that the relationship between the decimated and the original sequences becomes obvious.
It's just a rearrangement of the terms of the sum. As a simple example, take an infinite sum of numbers $a_r$:
$$S=\sum_{r=-\infty}^{\infty}a_r\tag{1}$$
Under certain conditions that we don't need to bother with now we can rearrange the sum and write it like
$$\begin{align}S&amp;=\ldots+a_{-2M}+a_{-M}+a_0+a_M+a_{2M}+\ldots\\&amp;\ldots+a_{-2M+1}+a_{-M+1}+a_1+a_{M+1}+a_{2M+1}+\ldots\\&amp;\vdots\\&amp;\ldots+a_{-2M+(M-1)}+a_{-M+(M-1)}+a_{0+(M-1)}+a_{M+(M-1)}+a_{2M+(M-1)}+\ldots\tag{2}\end{align}$$
with some arbitrarily chosen integer $M$. So we just start with element $a_0$ and add every $M^{th}$ element, then we move to element $a_1$ and add again every $M^{th}$ element, etc. If we do this $M$ times, we've added all elements, just like in the originial sum $(1)$.
Eq. $(2)$ can be written as
$$S=\sum_{i=0}^{M-1}\sum_{k=-\infty}^{\infty}a_{i+kM}\tag{3}$$
which means that we expressed the index $r$ as $r=i+kM$.