For 3-simplices (i.e. tetrahedra), I understand the basic idea behind the Pachner moves 1 $\leftrightarrow$ 4, which takes one tetrahedron and replaces it with four (or vice versa), and 2 $\leftrightarrow$ 3, which shifts between 2 and 3 tetrahedra. These moves can also be defined on the dual graph to the tetrahedron, where the graph has a 4-valent node corresponding to each 3-simplex, and each edge in the graph corresponds to a face of the tetrahedron.
However, I am curious specifically about performing this 2 $\leftrightarrow$ 3 move on an arbitrary 4-valent graph. Is there an ambiguity about which edges you put together at each of the new nodes, when you go in the 2 $\to$ 3 direction? Since you seem to lack information about whether a triangulation is dual to the graph, it appears to me you have six possible choices of how to join the edges up. Going the other direction, 3 $\to$ 2, also seems to have an ambiguity in how to group edges into two groups of three.
Yes, the ambiguity that you mention does exist. This is because the dual graphs do not completely represent a triangulation. Each triangulation has one dual graph; but each such dual graph can be created from many triangulations.
Self plug: If you want to see a modification of the dual graph representation that does not lose any information of this information, feel free to read through http://arxiv.org/abs/1412.2169. In this setting you can perform the $2 \leftrightarrow 3$ and $1 \leftrightarrow 4$ Pachner moves without any ambiguity.