Trapezoidal decomposition of a graph
In order to get valid paths in the free space from paths in the graph $G=(V,E)$ that is constructed via the trapezoidal decomposition, we require that all edges of $G$ lie inside the free space.
The segment between two centers of adjacent trapezoids is not always fully contained inside the free space. For example, consider a square and take a slightly smaller square with the same center as the obstacle. So, using only the nodes at the center of the trapezoid is not sufficient to find paths in the free space.
On the other hand, any segment between a pair of nodes centered at a vertical boundary of the same trapezoid is fully inside the free space, since trapezoids are convex. So, the graph $G'=(V',E')$ with as vertices the centers of the vertical boundaries with an edge between every pair of vertices that lies on the boundary of the same trapezoid lies fully within the free space.

So, why bother using the graph $G$ and not $G'$? One issue with $G'$ is that a trapezoid with $d$ adjacent trapezoids means $G'$ has the complete graph on $d$ nodes as a subgraph, which has $\Theta(d^2)$ edges. So in the worst case, finding a path in $G'$ takes at least $\Omega(n^2)$ time, where $n$ is the number of  in the decomposition. On the other hand, $G$ has at most $2$ edges for each vertical trapezoid boundary, so at most two per vertex of the obstacle boundary, which means it has $O(n)$ edges and we can find shortest paths in $O(n\log n)$ using Dijkstra.