I think that there's a fairly useless argument using Riemannian geometry. A PL manifold admits a compatible smooth structure (but note that the [smooth structure need not be unique][1]). Take a Riemannian metric on the manifold, and rescale it so that the curvature is nearly zero, and the injectivity radius is large everywhere. We may also assume that each point is contained in a large normal neighborhood, say for  some radius $R>>1$. Then take a net of points in the manifold, a maximal collection of points with distance $> \epsilon$ between every pair, $\epsilon <<1$. Then the $2\epsilon$ balls about these points cover the manifold. Now take the [Delaunay triangulation][2] with respect to this covering. Because the metric is nearly flat on the scale $2\epsilon$, this will give a triangulation of bounded degree. 

There are many details that need to be checked for this argument to work, but I expect that this is the sort of construction Cooper-Thurston may have had in mind. Maybe to convince oneself that this might work, the rescaled limit about any point for a Riemannian metric (the [asymptotic cone][3]) will be Euclidean, and the net will be locally finite, hence the Gromov-Hausdorff limits of such triangulations will be $\epsilon$-net Delaunay triangulations of Euclidean space, which have bounded degree. Each edge will be between vertices of distance at most $2\epsilon$, so the $3\epsilon$ ball about every vertex will contain the disjointly embedded epsilon balls about the vertex and its neighbors. Hence the degree ($+1$) will be at most $3^d$ where $d$ is the dimension. 


  [1]: https://mathoverflow.net/a/97019/1345
  [2]: https://en.wikipedia.org/wiki/Delaunay_triangulation#Properties
  [3]: https://en.wikipedia.org/wiki/Ultralimit#Asymptotic_cones