I think I somehow got confused with orientations of a $\Delta$-complex structure, so I wrote down what I think to confirm if they are correct. Thank you.
When we try to put a $\Delta$-complex structure on a 2-simplex, we cannot put the 1-simplex, the edges, into a cyclic orientations. This is because the restriction of $\sigma:\Delta_2\to X$ to its edges respect the canonical order. For example, $\Delta_2=[v_0, v_1, v_2]$. Its edges are obtained by ignoring one of the three vertices and the order is determined by the dropped vertices. The orientation of the edge dropping $v_i$ is $(-1)^i$. This means the orientation of the three edges cannot be cyclic. The figure on the left is a feasible choice for the $\Delta$-complex structure.
However, consider the dunce hat defining by gluing all the three edges in a cyclic pattern. In this case, we can only put $\Delta$-complex structure as in the right figure since all three edges are already glued in this order so actually there is only one edge. There is no needs to consider about ordering of edges. In this way, we find the simplicial homology group of the dunce hat is $H_0(X)=\mathbb{Z}$, $H_0(X)=\mathbb{Z}/3\mathbb{Z}$, $H_n(X)=0$ for $n\geq 2$.
In summary, my points are
-
For a usual 2-simplex, we need to respect the orientatiaon so the ordering of edges must not be cyclic.
-
However, if we glue the three edges in a pattern, we no longer need to consider the problem of orientation and just follow the pattern of gluing. In fact in this case there is only one edge.
Best Answer
That structure on the space $X$ that you have depicted to is not a $\Delta$-complex structure, as given in Hatcher's Algebraic Topology Section 2.1.
In order to have a $\Delta$-complex on $X$ with one 1-simplex $a$ as depicted and with one 2-simplex $U$ as depicted, you must choose a continuous function $\sigma_1 : \Delta^1 \to X$ corresponding to $a$, and another continuous function $\sigma_2 : \Delta^2 \to X$ corresponding to $U$, so that the restrictions of $\sigma_2$ to all three sides of $\Delta_2$ agree with $\sigma_1$.
More formally, denoting $\Delta^1 = [w_0,w_1]$ and $\Delta^2 = [v_0,v_1,v_2]$, each of the following three composed maps must equal the map $\sigma_1 : [w_0,w_1] \to X$: \begin{align*} [w_0,w_1] & \mapsto [v_0,v_1] \subset [v_0,v_1,v_2] \xrightarrow{\sigma_2} X \\ [w_0,w_1] & \mapsto [v_0,v_2] \subset [v_0,v_1,v_2] \xrightarrow{\sigma_2} X \\ [w_0,w_1] & \mapsto [v_1,v_2] \subset [v_0,v_1,v_2] \xrightarrow{\sigma_2} X \end{align*} But, as you have already said, this is not possible because the three sides of $\Delta_2 = [v_0,v_1,v_2]$, namely $[v_0,v_1]$ and $[v_0,v_2]$ and $[v_1,v_2]$, are not cyclically ordered as you have depicted.
By the way, I should warn you that the space $X$ you depicted in your linked picture is not the dunce hat. In the dunce hat as depicted here you will see that the edge identifications do not obey the cyclic orientation that you have depicted, but instead are oriented exactly as required for the definition of a $\Delta$-complex to hold, IF one picks the correct function $\sigma_2 : X \to X$