Shortest cycle separator for biconnected planar graphs
This doesn't seem to be possible, at least not without finding the right notion of "easy".

Consider a graph formed by taking a grid graph with $k$ rows and $l+1$ columns, and identifying the leftmost column of vertices with the rightmost column of vertices. Here, we have $n = kl$ vertices, and we're interested in $k \le l$. The graph is planar, and has treewidth at least $k$, as it has a $k\times k$ grid graph as a minor.

Additionally, a cyclic separator achieving constant $f$ in this graph seems to need $\Omega(l)$ vertices (where the constant depends on $f$). Imagine the graph drawn on a cylinder of circumference $l$ and height $k$, where the graph forms an equally spaced grid on the surface of the cylinder. Then if a cyclic separator has length less than $l$, it cannot "go around" the cylinder. Hence a cyclic separator gives a way to find a region in a $k\times l$ rectangle whose perimeter is the length of the separator, and whose area is at least a constant ($1-f$) fraction of the whole rectangle's area. I don't know the exact isoperimetric inequality for this, but it seems like such a region would need perimeter $\Omega(l)$ in order to have enough area and fit into the rectangle.

Thus we have a graph with treewidth $k$ and whose separators need to be of size $\Omega(l)$. Choosing parameters like $k = n^\epsilon$ and $l = n^{1-\epsilon}$, we thus see that a cyclic $(s,f)$-separator for constant $f$ needs to have $s = \Omega(n^{1-\epsilon})$, even for graphs with treewidth $n^\epsilon$.