= Solution
Write the wedge's <fundamental group> as the <free group> on $a,b$. Use the index-two subgroups
$$
H=\ker(a\mapsto1,b\mapsto0),\qquad H'=\ker(a\mapsto0,b\mapsto1),
$$
where each map has target $\mathbb Z/2$. Explicitly, the first cover has two vertices, two directed lifts of $a$ exchanging the vertices, and a $b$-loop at each vertex. The second interchanges the roles of $a,b$. Each vertex has exactly one incoming and one outgoing lift of each labelled edge, so these are genuine two-sheeted <graph> coverings. They are <connected>, and a chosen vertex gives a based cover.
The subgroups differ because $a$ belongs to $H'$ but not $H$, while $b$ belongs to $H$ but not $H'$. Hence \b[the covers have the same degree and are inequivalent].
\Image[/past-exam-of-the-topology-and-groups-and-algebraic-topology-course-of-the-university-of-oxford/2007-two-sheet-covers.png]
{title=Two inequivalent double covers of a two-circle wedge}
{description=In one cover the a-edges exchange the sheets and the b-edges are loops; in the other the roles are reversed. The left vertex is the chosen basepoint in each.}
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