Detect if a graph has a $k$ cycle in space complexity $O((\log k)^d)$ for fixed $d \geq1$
No, this is impossible for your parameters.  With $s$ bits of space, you can only visit at most $2^s$ vertices of the graph.  Now set $s = O((\log k)^d)$ and $k=(\log n)^2$ and it is clear that you cannot visit all of the graph, as $2^s = o(n)$.  Thus for any algorithm that uses only $O((\log k)^d)$ bits of space, there exist graphs where you fail to detect a $k$-cycle (e.g., because the $k$-cycle is in the unvisited part of the graph).

If $k=n^{\Theta(1)}$, and if your definition of a cycle allows repeated vertices to appear in the cycle, then an algorithm exists.  A simple approach is to guess a starting vertex $v_0$ and store it (which requires $\log n = O(\log k)$ bits); then, non-deterministically guess a path of length $k$, counting up to $k$ as you go (which requires $\log k$ bits), checking that none of the intermediate vertices are equal to $v_0$ and the final vertex is equal to $v_0$.
If $k=n^{\Theta(1)}$, and if your definition of a cycle does not allow repeated vertices to appear in the cycle, then I don't know whether an algorithm exists or not.