This is a good exercise. As far as I understand the OP, the situation is as follows. Given a deterministic non-negative function $\phi$ on the interval $[0,1]$ with $\phi(t)=0\iff t \in\{0,1\}$ and a sequence of random variables $Z_n$, one defines $$ \zeta(t)=\phi(t-n) \cdot Z_n \;, \qquad n\le t\le n+1 \;. $$ The question is when $\zeta(t)$ is Markov, and the answer to this question is pretty obvious: if and only if the variables $Z_n$ are independent (look at the Markov condition at integer times).

In the original question $$ \phi(t)=t(1-t) $$ and $$ Z_n=(\xi_1^2+\dots+\xi_n^2)(\xi_1^2+\dots+\xi_{n+1}^2) \tag{$\diamond$} \;, $$ which are pretty obviously not independent unless $\xi_i^2$ are a.s. constant.

This is a good exercise. As far as I understand the OP, the situation is as follows. Given a deterministic non-negative function $\phi$ on the interval $[0,1]$ with $\phi(t)=0\iff t \in\{0,1\}$ and a sequence of random variables $Z_n$, one defines $$ \zeta(t)=\phi(t-n) \cdot Z_n \;, \qquad n\le t\le n+1 \;. $$ The question is when $\zeta(t)$ is Markov, and the answer to this question is pretty obvious: if and only if the variables $Z_n$ are independent (look at the Markov condition at integer times).

In the original question $$ \phi(t)=t(1-t) $$ and $$ Z_n=(\xi_1^2+\dots+\xi_n^2)(\xi_1^2+\dots+\xi_{n+1}^2) \;, $$ which are pretty obviously not independent unless $\xi_i^2$ are a.s. constant.