The point is that $A\subset [1,N/k)$ naturally gives a subset of $\mathbb{Z}/N\mathbb{Z}$ of the same size under the obvious identification, and $A$ being concentrated in a short interval prevents any non-trivial k-APS being created where there were none before, since there is no 'wrap-around'.
More formally, we can argue as follows.
The main trick is to consider the natural bijection $\phi:\mathbb{Z}/N\mathbb{Z}\to \{1,\ldots,N\}$ (so $x\equiv \phi(x)\pmod{N}$) and note that if $\phi(a)+\phi(b)<N$ then $\phi(a+b)=\phi(a)+\phi(b)$. More generally, if $\phi(a_1)+\cdots +\phi(a_r)< N$ then
$$ \phi(a_1+\cdots+a_r)=\phi(a_1)+\cdots+\phi(a_r).$$
Let $A\subset [1,N/k)$ have no k-APs, and let $A'=\phi^{-1}(A)$, and suppose that there is a k-AP $x,\ldots,x+(k-1)d\in A'$ with $d\neq 0$. Without loss of generality, $\phi(d)\leq N/2$ (or else consider the k-AP with $x$ replaced by $x+(k-1)d$ and $d$ replaced by $-d$). I claim that $\phi(x),\ldots,\phi(x+(k-1)d)$ is also a k-AP - then we have a non-trivial k-AP in A, contradiction, and hence $A'$ is also k-AP free, and so $r_k([1,N/k))\leq r_k(\mathbb{Z}/N\mathbb{Z})$.
Indeed, we will show that $\phi(x),\ldots,\phi(x+(k-1)d)$ forms a k-AP by explicitly showing that $\phi(x+id)=\phi(x)+i\phi(d)$ for $0\leq i<k$. By the above additive property of $\phi$, it suffices to show that $\phi(x),\phi(d)<N/k$. This is trivial for $\phi(x)$ (since $\phi(x)\in A\subset [1,N/k)$).
Furthermore, we have $d\equiv \phi(x+d)-\phi(x)\pmod{N}$. If $\phi(x+d)>\phi(x)$ then $\phi(d)=\phi(x+d)-\phi(x)<N/k$ (using $\phi(x),\phi(x+d)\in A\subset [1,N/k)$). If $\phi(x)>\phi(x+d)$ then $\phi(d)=N+\phi(x+d)-\phi(x)>N-N/k>N/2$, which is a contradiction to our assumption that $\phi(d)\leq N/2$.
