Geometric Interpretation of S3 My impression was that the symmetric group $S_3$ acts on the vertices of a labeled triangle. However, I am not sure this is the case anymore, because of the following. (The triangle is labeled as below:)

If $S_3$ acts on a triangle, it would be nice if (123) is counterclockwise rotation, and (23) is a flip over the vertical axis. (I was under the impression that (123) means putting the vertex labeled 1 in the place of the vertex labeled 2, the vertex labeled 2 in the place of the vertex labeled 3, and the vertex labeled 3 in the place of the vertex labeled 1.)
(123)(23)—multiplying first the flip transformation and then the rotation transformation gives (12), which is a flip over the left axis. With an actual triangle, flipping over the vertical axis and rotating counterclockwise gives you (12) again. This is consistent with the permutation notation result.
However, if you interpret the action of $S_3$ as acting on the numbered vertices, as (23) exchanging number 2 and number 3, doing the sequence above, leads to a different result: you get (13). (The counterclockwise rotation becomes a clockwise one.)
If $S_3$ acts on the places, 1, 2 and 3, this seems to make more sense, and the action would match the observed symmetry of the triangle. However, how do you think about this? When I have been introduced to it they have said (or at least I came away with the impression) that the group was acting on the vertices.
If you are trying to teach this and transition from the physical model to the cycle notation, you can just emphasize this is about moving the places. However, to operate with the cycle notation, you need to think about numbers and not places. Has anyone found this to be a problem for students, or am I thinking about it wrong?
 A: You seem to be using two different reflections in your counterargument. First you say that $(123)(12)=(13)$, which is true, but when you "actually do the flip then rotate the triangle counterclockwise" you are applying $(23)$ then $(123)$, which is $(12)$. However $(123)(23)=(12)(23)(23)=(12)$, and this is the product you should have computed originally, so this is not a counterexample.
Indeed, there are no "holes" in the interpretation, and $S_3=D_3$ is the symmetry group of a triangle. This action is equivalent to permutation of the vertices.
Edit: The issue in your edit arises from the fact that it matters which side the group acts on. If you are permuting the numbers by moving the number currently at a position to a new position, this is a right action of $S_3$. This explains why the result is not the same. The permutation on the left in the product is applied first.
A: 
If instead it acts on the places, 1, 2 and 3, this seems to make more sense, and the action would match the observed symmetry of the triangle. However, it does not solve the problem that multiplying counterclockwise rotation with a vertical flip is not the same as multiplying the elements you would like to say represent these transformations in S3.

This is incorrect. The numbers $1,2,3$ are indeed static positions and the vertices move between these points, but there is no inconsistency between multiplying the permutation elements and composing their actions on the triangle.
Label the vertices in the initial configuration such that $A,B,C$ are at $1,2,3$, respectively, so $ABC$ for short. The reflection corresponding to the action of $(12)$ on $ABC$ gives $BAC$; similarly, $(123)$ acts on $BAC$ to give $CBA$. It's clear that composing these permutations $(123)(12)$ is equivalent to the permutation $(13)$, which acts on $ABC$ to yield $CBA$ as you noted.
