If $k$ is not a power of $2$, then $k$ has an odd factor $\geq 3$, so $k = 2^j \cdot m$ for some $j \geq 0$ and odd $m \geq 3$. In the Sharkovsky ordering, $3 \cdot 2^j \triangleleft k$ (since $3$ precedes all other odd numbers and all their multiples). By Sharkovsky's theorem, $f$ has a $3 \cdot 2^j$-cycle, which means $f^{2^j}$ has a $3$-cycle. By the period-three-implies-chaos theorem applied to $f^{2^j}$, the map $(f^{2^j})^2 = f^{2^{j+1}}$ has a horseshoe. Hence $f$ is Glendinning-chaotic.