 Solution

Put $r=n-1$. Take $r$ spheres, each mapping by the double cover onto ${\mathbb{RP}}^2$, and write their antipodal sheet points as $a_j,b_j$ for $0\le j<r$. Add an oriented $x$ edge from $a_j$ to $a_{j+1}$, with indices modulo $r$, and add an $x$ loop at each $b_j$. Every sheet point has the required two circle branches, so the resulting space is a $2r$-sheeted cover. It is connected because the $a_j$ edges form a cycle and each sphere connects its own two sheet points.

Choose a connecting arc from $a_j$ to $b_j$ in each sphere. Collapsing these arcs leaves the spheres simply connected, and leaves a graph with $r$ vertices, one cyclic edge at each step through the $a_j$, and one additional loop per vertex. This graph has $2r$ edges and rank $2r-r+1=r+1=n$. The sphere summands add no fundamental-group generators. Thus the cover has free fundamental group of rank $n$.

Based at $a_0$, choose the first $r-1$ cyclic $x$ edges as a spanning tree. The closing cycle projects to $x^r$. To traverse the loop at $b_j$, travel along the tree by $x^j$, use the sphere connector projecting to $y$, traverse its $x$ loop, and return. The resulting free basis in the base group isThere are exactly $n$ elements. For $n=2$ this is the previous pair $x,yxy$. Again injectivity of a covering's induced map makes these actual subgroups with the claimed free bases, rather than just quotient presentations.

This realizes Free subgroups of every finite rank in the free product of Z and C2.