We usually choose the fundamental domain of full modular group $\Gamma$ to be $\mathcal{F} = \{z\in \mathbb{H} \, | \, |z|>1, \, |\text{Re}(z)|<\frac{1}{2} \}$. Using the change of variables $z \mapsto \frac{z-i}{z+i}$ the upper half plane $\mathbb{H}$ will be mapped to Poincaré disc $D = \{ z \, | \, |z|<1 \}$.
How does the fundamental domain $\mathcal{F}$ look in this disc $D$? It seems that atleast the cusp at infinity is mapped to $1$. And maybe by choosing different fundamental domain all the cusps are mapped to the boundary of the disk?
Additionally, if we define a cusp to be an element of $\mathbb{R} \cup \infty$ such that it is fixed by parabolic element of $\Gamma$ (Non-trivial element with trace $2$), how can we see that this definition is satisfied by those boundary points?
It's an isosceles ideal triangle with angles $\frac{\pi}{3}$, $\frac{\pi}{3}$, $0$. You can't say anything beyond that because of the large isometry group. To find the other two vertices, apply your map to the cube root and the sixth root of $1$, respectively. All such triangles are isometric to each other; in that sense their position is irrelevant.