Wikipedia has the answer in the case of the Poincaré disk model. When the point $\mathbf{v}$ is translated to the origin, then the point $\mathbf{x}$ is translated to
$$\frac{(1 + 2\mathbf{v} \cdot \mathbf{x} + |\mathbf{x}|^2)\mathbf{v} + (1-|\mathbf{v}|^2)\mathbf{x}}{1 + 2\mathbf{v} \cdot \mathbf{x} + |\mathbf{v}|^2|\mathbf{x}|^2}.$$
What is the equivalent for the hyperboloid model?
I've tried a variety of solutions and I haven't found any that are undeniably right. I've also looked for solutions and didn't find any that were concrete like the example given above.
First you need the Lorentz inner product $$\underbrace{( t_0, x_0, y_0)}_{\vec v^T} \underbrace{\pmatrix{-1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1}}_J \underbrace{\pmatrix{t_1 \\ x_1 \\ y_1}}_{\vec w} = -t_0 t_1 + x_0 x_1 + x_2 y_2 $$ which I'll write in shorthand as $$\vec v^T \, J \, \vec w $$ The hyperboloid model $\mathbb{H}$ is simply the set of column vectors $\vec v$ that are solutions of the equation $$\vec v^T \, J \, \vec v = -1, \, t_0 > 0 $$ in other words $\vec v^T = (t,x,y) \in \mathbb{H}$ if and only if $-t^2 + x^2 + y^2 = -1$.
Next you need Lorentz transformations, the $3 \times 3$ matrices $M$ having the property that $$M^T J M = J $$ These matrices form a group under matrix multiplication, that group is denoted $SO(1,2)$. Each of these matrices takes $\mathbb{H}$ to itself, and is an isometry of $\mathbb{H}$. This is the full isometry group of $\mathbb{H}$ in the hyperboloid model.
Once you have those things, your problem has an easy solution. Given the coordinates of column vectors $\vec v$, $\vec w \in \mathbb{H}$, you must solve for the unknown coordinates of a Lorentz transformation $M$ such that $$M \vec v = \vec w $$ Now your problem is just a big fat set of linear transformations to be solved, expressed in the unknown coordinates of $M$. You can get the final answer to your problem by setting $\vec w$ to be the zero vector.