Shortest path with positive edge cycle
They are just saying that if $P$ is a shortest path, in $P$ there are not positive cost cycles, since removing the cycle would led to a smaller cost (and so to a better path).

Consider your example and suppose we want the shortest path beween $s$ and $v_k$, let $P=(s,v_1,\color{red}{v_i},\color{red}{v_{i+1}},\color{red}{v_{j-1}},\color{red}{v_i},v_{i+1},v_{j-1},v_{j+1},v_k)$.

Removing the cycle from $P$ we obtain $P'=(s,v_1,v_i,v_{i+1},v_{j-1},v_{j+1},v_k)$ where $cost(P')=cost(P)-6$, so to a better path in terms of cost.