In a Riemannian manifold consider two compact smooth submanifolds $S$, $S^\prime$ that intersect transversely. It seems intuitively obvious that for a sufficiently small number $r$, the union of $r$-neighborhoods of $S$ and $S^\prime$ is a regular neighborhood of $S\cup S^\prime$ in some triangulation of the ambient manifold. Is this written somewhere?
By the way, a standard reference for regular neighborhoods is this paper by Marshall Cohen, published in Trans. Amer. Math. Soc. 136, 1969 189--229.
EDIT: it may be worth discussing why any small $r$-neighborhood $N_r(S)$ of a smooth compact submanifold $S$ is a regular neighborhood. This is written on page 527 of [M. Hirsch, Smooth regular neighborhoods., Ann. of Math. (2) 76 1962 524--530]. First, one uses results of J. Whitehead to show that $N_r(S)$ is a subcomplex in some smooth triangulation. By choosing $r$ small enough we arrange that $N_r(S)$ is inside the second derived neighborhood $R$ of $S$, i.e. the star of $S$ in the second barycentric subdivision. The second derived neighborhood is always a regular neighborhood. Furthermore, Hirsch says $R$ is the mapping cylinder of a retraction $\partial R\to S$, and each radial segment in the cylinder is transverse both to $S$ and to $\partial R$. This transversality implies that for small $r$ removing the interior of $N_r(S)$ from $R$ gives $\partial R\times I$, so that $N_r$ is also a regular neighborhood of $S$.
The same argument may work for $S\cup S^\prime$. In fact it is believable that the second derived neighborhood $R$ of any subcomplex $K$ can be written as a mapping cylinder whose radial segments are PL-transverse to $\partial R$ and $K$. If true, we just need to see that the segments are also transverse to the boundary of the union of $r$-neighborhoods of $S$ and $S^\prime$.