Before making the identifications, consider the following triangulations in the Euclidean plane, and the simplicial map given by the labelling:
Call this map $f\colon X \to Y$, where the numbered vertices are mapped as $f(v_i) = v_i'$ for $1\leq i\leq 7$, each edge $\overline{v_i v_j}$ is mapped as $f(\alpha v_i + \beta v_j) = \alpha v_i' + \beta v_j'$ for any $\alpha + \beta = 1$ with $\alpha, \beta \geq 0$ and each triangle $\triangle v_i v_j v_k$ is mapped as $f(\alpha v_i + \beta v_j + \gamma v_k) = \alpha v_i' + \beta v_j' + \gamma v_k'$ for $\alpha + \beta + \gamma = 1$ with $\alpha, \beta, \gamma \geq 0$. This is just mapping barycentric coordinates in each triangle in $X$ to the corresponding barycentric coordinates of the corresponding triangle in $Y$. This is continuous by the pasting lemma, and each $y\in Y$ has singleton preimage $f^{-1}(y)$, except $f(v_1) = v_1' = v_5' = f(v_5)$.
Now your arrows define some equivalence relation $\sim$ on the left diagram and some equivalence relation $\sim'$ on the right diagram. We have quotient maps $\pi \colon X \to X/\sim$ and $\pi'\colon Y \to Y/\sim'$.
Define $F \colon X/\sim \to Y/\sim'$ by $F(\pi(x)) := \pi'(f(x))$. To show that this is well-defined, we need to show that if $\pi(x_1) = \pi(x_2)$ then $\pi'(f(x_1)) = \pi'(f(x_2))$. Said differently, we need to show that if $x_1 \sim x_2$ then $f(x_1) \sim' f(x_2)$. This is easily verified: however you're defining what such a diagram even is, every identification made on the left is also an identification made between corresponding points on the right. By the universal property of the quotient topology $X/\sim$, we know $F$ is continuous ($F$ is characterized by $F \circ \pi = \pi' \circ f$).
$F$ is surjective because $f$ is surjective: each $\pi'(y) \in Y/\sim'$ has some $x\in f^{-1}(y)$, so $\pi'(y) = \pi'(f(x)) = F(\pi(x)) \in \operatorname{Image}(F)$.
We can also show $F$ is injective. Suppose $F(\pi(x_1))=F(\pi(x_2))$. So $\pi'(f(x_1))=\pi'(f(x_2))$, and so $f(x_1) \sim' f(x_2)$. We need to show that $\pi(x_1) = \pi(x_2)$, i.e., that $x_1 \sim x_2$.
- If $f(x_1) \in Y^\circ$ then since $f(x_1) \sim' f(x_2)$ we must have that $f(x_2) \in Y^\circ$ as well. Since $\sim'$ is trivial on $Y^\circ$ we must have $f(x_1)=f(x_2)$. And since $f$ is injective on $f^{-1}(Y^\circ) = X^\circ$, we must then have $x_1 = x_2$.
- If $f(x_1) \in \{v_1', \dots, v_5'\}$ then since $f(x_1) \sim' f(x_2)$ we must have that $f(x_2) \in \{v_1',\dots, v_5'\}$ as well. But then $x_1$ and $x_2$ must belong to $\{v_1, \dots, v_5\}$, all of which are identified in $X/\sim$.
- If $f(x_1)$ belongs your open edge "$c$" in $Y$, then since $\sim'$ is trivial on the open edge $c$, we know $f(x_1) = f(x_2)$, and since $f$ restricts to a bijection between the open edge $c$ in $X$ and that of $Y$, we conclude that $x_1=x_2$.
- If $f(x_1)$ belongs to some open edge labelled $a$ or $b$ in $Y$, then $f(x_2)$ must be one of the two points related by $\sim'$. Each has a unique preimage $x_2$ under $f$, but both possible preimages are related by $\sim$ by construction, so $x_1\sim x_2$. In other words, on the open edges labelled $a$ or $b$, $f$ restricts to a bijection that maps $\sim$ to $\sim'$.
So we can conclude that $F$ is a bijection. The domain $X/\sim$ is compact because $X$ is compact, so if we can show that $Y$ is Hausdorff then we can conclude that $F$ is a homeomorphism. $Y$ is indeed Hausdorff by more casework:
- Points in $Y^\circ/\sim'$ are separated by open sets because the corresponding points $Y^\circ$ are separated by open sets, and $\pi'$ is open on $Y^\circ$.
- Any point in $Y^\circ/\sim'$ is separated from any point in $\partial Y/\sim'$ because we can take a small closed neighborhood, away from $\partial Y$, of a point in $Y^\circ$, along with an open subneighborhood, and then these map to such closed and open neighborhoods of in $Y/\sim'$, so the complement of the closed neighborhood is an open neighborhood of $\partial Y/\sim'$ disjoint from the open neighborhood of the given point.
- Points in differently-labelled open edges of $\partial Y$ can be separated by open sets in $\partial Y/\sim'$ because we can take thin triangular neighborhoods around each edge in $Y$, which will get mapped to open neighborhoods of each open edge in $Y/\sim'$
- Points within an open edge can be separated by open sets because if one point has parameter $t_1\in[0,1]$ along the edge and the other has parameter $t_2$, then take $\epsilon = \min\left(\frac{|t_1-t_2|}{2}, t_1, t_2, 1-t_1, 1-t_2\right)$ and form small $\epsilon$ balls around the two points, intersected with the thin triangles from before just to be safe. Since the parameterizations agree in the quotient, these are disjoint open sets.
- Vertices in $Y\sim'$ can be separated from any other point by choosing sufficiently small balls around all vertices in $Y$ at once.
You can fill in as many more details as your heart desires. Much of this could probably also be packaged into more general purpose lemmas depending on exactly what your use case is.
