The Dirichlet fundamental domain for a Fuchsian group $\Gamma$ acting freely on the hyperbolic plane $\mathbb H^2$ is a particular type of fundamental domain in hyperbolic geometry. For $z_0 \in \mathbb H^2$ we define the Dirichlet domain for $\Gamma$ as $$P:=\bigcap_{g \in \Gamma}\{z \in \mathbb H^2 : d(z,z_0)\leq d(gz,z_0)\}$$
and it indeed tessellates $\mathbb H^2$, meaning that $$\mathbb H^2 = \bigcup_{g \in \Gamma}g P\quad \text{and} \quad g\, \mathrm{Interior}(P) \cap \mathrm{Interior}(P) = \emptyset \text{ for } g \neq \mathrm{id}.$$
Not necessarily the fundamental domain you are working on is a Dirichlet one. The most common type of fundamental domain comes from Poincaré's theorem for fundamental polygons: consider a convex geodesic polygon
$P$ with finite sides $S$, a bijection $\overline \bullet :S \mapsto \overline S$ that maps $s \to \overline s$ satisfying $\overline{\overline{s}} = s$, and orientation-preserving isometries $I_s:\mathbb H^2 \to \mathbb H^2$ for each $s \in S$. We also assume that $I_s(s)=\overline s$, $I_s^{-1} = I_{\overline s}$ and $I_s P \cap P = \overline s$. In other words, we have a polygon $P$, and for each side $s$ we associate an isometry $I_s$ such that $I_s$ maps $P$ to an adjacent polygon $I_sP$ sharing $\overline s$ with $P$.
Under these conditions, we can talk about the cycle of vertices. For each vertex $v_1$ we have a cycle constructed in the following way. Take $s_1$ to an adjacent side to $v_1$. We have that $v_2:=I_{s_1}v_1$ has $\overline s_1$ as an adjacent side. We define $s_2$ as the other adjacent side. Then we do the same for $v_2$ and $s_2$, we have $v_3:=I_{s_2} v_2$ and $I_{s_2}s_2$. We define $s_3$ to be the other side adjacent to $v_3$, and so on. Following this procedure we eventually obtain $v_{n+1}=v_1$ again and $s_{n+1}=s_1$. Thus, we obtain the cycle $$(v_1,s_1)\to (v_2,s_2) \to (v_3,s_3) \to \cdots \to (v_n,s_n) \to (v_1,s_1).$$
The idea is to impose that the sum of inner angles on the vertex $v_1, \ldots,v_n$ is $2\pi/m$ for some positive integer $m$. If for all vertices cycle this condition is met, then the group generated by the isometries $I_s$ tessellates $\mathbb H^2$ with fundamental domain $P$.
What all this means is the following. Assume for simplicity that $m=1$. Then, around $v_1$ we have the polygons $$P,\quad I_{s_1}^{-1}P,\quad I_{s_1}^{-1}I_{s_2}^{-1}P,\quad \ldots, \quad I_{s_1}^{-1}I_{s_2}^{-1}\cdots I_{s_{n-1}}^{-1}P.$$ tessellating around $v_1$. The idea is that tessellation around vertices implies the tessellation of the whole hyperbolic plane. Additionally, by studying the tesselation around each vertex we obtain $I_{s_n} \cdots I_{s_2}I_{s_1} =\mathrm{id}$ (and $I_{s_n} \cdots I_{s_2}I_{s_1}$ is a $\pm 2\pi/m$ rotation centered in $v_1$ for $m> 1$).
Now, about your particular problem. We have the polygon you described and we need isometries that generate the tesselation. Define $v_4 = R_1 v_1$ and consider the isometries $I_2 = R_1 R_3$ and $I_3 = R_2 R_1$. The isometry $I_2$ is the $-2\pi/m_2$ rotation centered at $v_2$ and $I_3$ the $-2\pi/m_3$ rotation centered at $v_3$. Note that $I_2(v_2v_1) = v_2v_4$ and $I_3(v_3v_4) = v_3v_1$. The cycle of vertices:
- for vertex $v_2$: $$(v_2,v_2v_1)\to(v_2,v_2v_1);$$
- for vertex $v_3$: $$(v_3,v_3v_4)\to(v_3,v_3v_4);$$
- for vertex $v_1$: $$(v_1,v_1v_2)\to(v_4,v_4v_3)\to (v_1,v_1v_2);$$
- for vertex $v_4$: $$(v_4,v_4v_3)\to(v_1,v_1v_2)\to (v_4,v_4v_3).$$
The cycle for $v_2$ has only $v_2$ as vertex and the sum of angles is $2\pi/m_2$ and $I_2^{m_2}= \mathrm{id}$. The same goes for $v_3$. For $v_1$, the cycle has two points, $v_1,v_4$, and the sum of angles is $2\pi/m_1$. Additionally, $(I_3I_2)^{m_1} = \mathrm{id}$ (the isometry $I_1:=(I_3I_2)^{-1}=R_3R_2$ is a rotation centered at $v_1$ and angle $-2\pi/m_1$). Thus, we obtain that your quadrilateral tesselates under the action of the group
$$\Gamma := \langle I_1,I_2,I_3: I_j^{m_j} = \mathrm{id} \quad \text{and}\quad I_3I_2I_1 = \mathrm{id} \rangle.$$
For details about this group, see Von Dyck groups.
