To get some intuition behind the double suspension theorem, you can try three pages with a lot of pictures in Ferry's notes (starting with p.166, in Chapter 26). He gives a rough sketch of proof in the case of one particular homology sphere, the one for which the theorem was first proved by Edwards.
The double suspension theorem boils down to showing that a certain non-manifold (namely, the single suspension over a homology sphere) becomes a manifold when multiplied by $\Bbb R$. When Milnor conjectured the double suspension theorem in early 60s, he must have been aware of the existence of other non-manifolds with this property (found earlier by Bing). It is fortunate that they also exist in a lower dimension so it's easier to visualize what's going on. One such example is $(S^3/W)\times\Bbb R\cong S^3\times\Bbb R$, where $W$ is the Whitehead continuum, and there is a rather explicit construction of this homeomorphism in Ferry's Chapter 4 (p.15).
Added later: Another example is $(M/D)\times\Bbb R\cong M\times\Bbb R$, where $D$ is a wild copy of the $n$-disk contained in the interior of the manifold $M$. The case $n=2$ is actually used the above-mentioned proof of the double suspension theorem in Ferry's notes, but is not proved there; a proof of the case $n=1$ with some pictures can be found in the Daverman-Venema book, Section 2.6.
To address the specific question about neighborhoods, the open star (in the original double-suspension triangulation) of any vertex in the suspension circle is homeomorphic to $\Bbb R^5$ $-$ at least in the case of one particular homology sphere, the boundary of the Mazur manifold $W$. Indeed, the closed star of this vertex is the suspension over $cone(\partial W)$. As explained in Ferry's notes, $cone(\partial W)\times\Bbb R$ is homeomorphic to $W\times\Bbb R$. Hence the open star of the vertex is homeomorphic to $(W\setminus\partial W)\times\Bbb R$. The latter can be identified with the interior of $W\times I$. But it is easy to see that $W\times I$ is homeomorphic to the $5$-ball. So its interior is homeomorphic to $\Bbb R^5$.
