I would like help identifying the rigorous classification of this 'surface' geometry based on my interpretation of 3D models.

This is not an answer, but an extended comment. It might help understand the situation better.

OP has three triangles, $ABD$ (blue), $BDC$ (red), and $BDE$ (green). In the following picture, the vertices are labeled using letters, and edges with $1$, $2$, or $3$: The "real" side of the triangle is the one where the edges are numbered in counterclockwise order.

That is, $ABD$ (blue) and $BDE$ (green) have their real sides towards us, while $BDC$ (red) has its real side away from us.

If we consider the two-dimensional (and "one-sided") surface on those three "one-sided triangles", the geometry within each triangle is strictly Euclidean, but there are discontinuities at the edges of the 3D triangles. If we draw a directed graph of how a point on one triangle can pass to another triangle, by walking over an edge, we get:

The problem is that a 2D point in the $BDE$ (green) triangle traveling to the edge $1$ experiences a discontinuity at the edge. When it passes to the surface on the $ABD$ (blue) triangle, suddenly the entire halfspace (half-plane) behind it snaps to the $BDC$ (red) triangle!

If there were no such discontinuities, the geometry on the surface could be considered Euclidean.

A particular issue is that it is undefined which half-plane the 2D point observes, if it travels from the surface on the $BDE$ (green) triangle to the edge $1$, and then along the edge $1$: we cannot mathematically determine it based on the location of the point alone.

In computer programs, this problem is avoided by always having the point exist on the surface of exactly one triangle, and only changing to another triangle when the point passes over the edge. The discontinuous "snapping" happens only when the point goes outside the current triangle, onto a new one.

We can drop a dimension, since you can assume that locally the structure is the product of $\Bbb{R}$ and another geometric structure that I will describe.

This structure is a topological space together with a continuous action of the semigroup $\Bbb{R}_+$.

It is very similar to a switch in a train track structure. There are two semigroup actions of $\Bbb{R}_+=\{x\in\Bbb{R}$, s.t.\ $x\ge 0\}$, corresponding to moving forward and backwards. There can be no group action of $\Bbb{R}$, since that would imply time reversibility.

Consider three copies of the half line, blue, red and green, so the set is $S=\{b,r,g\}\times \Bbb{R}_+$.

Then the first semigroup action is given by $\rho_+:S\times \Bbb{R}_+\to S$ defined as \begin{eqnarray*} \rho_+((b,x),t)&=&\left\{ \begin{array}{ccc} (b,x-t)&if& t\le x\\ (g,t-x)&if& t>x \end{array}\right.\\ &&\\ \rho_+((r,x),t)&=&\left\{ \begin{array}{ccc} (r,x-t)&if& t\le x\\ (g,t-x)&if& t>x \end{array}\right.\\ &&\\ \rho_+((g,x),t)&=&(g,x+t)\\ \end{eqnarray*} and the second semigroup action is given by $\rho_-:S\times \Bbb{R}_+\to S$ defined as \begin{eqnarray*} \rho_-((b,x),t)&=&(b,x+t)\\ &&\\ \rho_-((r,x),t)&=&(r,x+t)\\ &&\\ \rho_+((g,x),t)&=&\left\{ \begin{array}{ccc} (g,x-t)&if& t\le x\\ (r,t-x)&if& t>x \end{array}\right.\\ \end{eqnarray*} The interpretation in terms of the structure described in the OP is the following: On the bottom you have the blue half line on the left and the green half line on the right, and the red half line is standing vertical on the joint. However, the joint is represented by the three different endpoints. The interpretation of the actions of $\Bbb{R}_+$ is that when you travel away from the joint, nothing special happens, but if you enter the joint from the blue half line, you don't see the red line and continue on the green line. If you came from the red line, you also continue on the green line, but if you come from the green line, the you must continue on the red line. Note that the actions $\rho_+$ and $\rho_-$ of $\Bbb{R}_+$ are not compatible, so they do not induce an action of $\Bbb{R}$.

You can also provide $S$ with a topology in order to make the action continuous: Take as basis for the topology all open intervals in $\Bbb{R}_+$ for each of the three halflines, add the open-closed-intevals on the blue line of the form $(b,[0,x))$ (containing the endpoint) and add the unions of red and green open-closed-intevals $(r,[0,x))\cup (g,[0,y))$. With other words, if an open set contains the blue endpoint it has to contain some blue end-interval, if it contains the red endpoint, it has to contain some red end-interval AND some green end-interval, and if it contains the green endpoint, it has to contain some green end-interval AND some red end-interval.

With this topology both actions are continuous.

The topology is not Hausdorff, in particular it comes not from a metric structure (but there is a pseudometric structure with the distance between the endpoints being zero) the point $(b,0)$ is closed, but the points $(r,0)$ and $(g,0)$ are not closed (the union of both points $\{(r,0),(g,0)\}$ is closed.

If you take the quotient space and collapse the three endpoints (identifying them as only one point), then the action is no longer well defined.

In general topological spaces with continuous actions of $\Bbb{R}$ which are also called flows, is very well studied, for example in the theory of differential equations. However, if you only have a semigroup action, or two actions that are not symmetric, there is very few literature.

If you now return to dimension two, you are dealing with oriented geodesics, I didn't find anything specifically referring to your particular problem, but Thurston has worked in related areas.

${Edit:}$

In the 2dim case you would have the semigroup $\Bbb{R}_+\times\Bbb{R}$ acting on three halfplanes (with two different actions).

$\newcommand{\Reals}{\mathbf{R}}$This may not be an answer, but is too lengthy for a comment.

For simplicity, let's push the dimension down by one. Consider the plane $\Reals^{2}$ with the "half-interval cut" $C = \{0\} \times (-1, 1)$, and assume:

1. The map $\gamma_{+}(t) = (t, 0)$, with $t$ real, is a continuous path. (Light can travel left-to-right through the cut.)

2. The map $\gamma_{-}(t) = (-t, 0)$, with $t \leq 0$, is a continuous path.

3. There exists no $a > 0$ such that $\gamma_{-}(t) = (-t, 0)$, with $t \leq a$, is a continuous path. (Light traveling right-to-left through $C$ "hits $C$ and stops/is absorbed/ceases to exist".)

I read the question as

How are we to describe such a geometric structure using the language/concepts and machinery of topology?

If the preceding properties capture the intent of a "half-triangle", one seemingly insurmountable obstacle to modeling with a topological space is: The map $\gamma_{-}$ is formally the composition of the continuous "time reversal" map $t \mapsto -t$ with the continuous map $\gamma_{+}$, but there exists no neighborhood of $0$ on which $\gamma_{-}$ is continuous (or even defined).

This violates the fact that in topology, a composition of continuous maps is continuous.