From Rotman's Algebraic Topology:
If $X$ is a polyhedron and $x \in X$, there exists a triangulation $(K,h)$ of $X$ with $x = h(v)$ for some vertex $v$ of $K$.
I'm having difficulty figuring out how to work this out, or even understand how it's possible for a simple example take $K = \Delta^2$ and $X = D^2$. How would you construct a homeomorphism if $p_0$, a vertex of $\Delta^2$, is mapped to the center of $D^2$?
Any suggestions?
In your example you cannot take $K = \Delta^2$, you have to refine it, in other words subdivide the simplices that appear. I hope the following picture is enough to answer your question (the little cross is $x$):
Essentially, let's say that you have a triangulation but $x$ is not already a vertex. Then $x$ is in the interior of some simplex $\sigma$, so you can subdivide $\sigma$ to get a new triangulation with more simplices such that $x$ is a simplex. However $\sigma$ is not necessarily of maximum dimension (consider the case where $x$ is in the boundary of $D^2$ but not at one of the vertices of the triangle, for example), so you may need to subdivide the higher-dimensional simplices that touch $\sigma$, then subdivide the higher-dimensional simplex that touch them, and so on: