I read that Moebius transformations are natural translation in hyperbolic plane
Yes, Möbius transformations which preserve the unit circle are exactly the orientation-preserving isometries of the disk model. Möbius transformations with two fixed points on the unit circle are the translations, with the fixed points the “endpoints” of the axis of translation. Contrary to Euclidean translations, a hyperbolic translation has a single designated axis, for lack of parallels.
I don't see how to choose a unique Moebius translation
What you wrote is not enough to make this unique. One way to choose a Möbius transformation for your scenario would be to find the translation which has the line from $x_0$ to $F(x_0)$ as it's axis of translation. This is an additional requirement which allows for a unique solution.
A Möbius transformation is uniquely defined by three points and their images. So you could start by drawing a line through $x_0$ and $F(x_0)$. This should be a hyperbolic line, but since it passes through the center, it's a Euclidean line as well. Intersect that line with the unit circle. Both these ideal points will be fixed points of the translation, while $x_0$ maps to $F(x_0)$. So here you have three pairs of points, and can compute the Möbius transformation. Ask a separate question if you need instructions on finding a Möbius transformation given three points, or adapt this answer of mine to $\mathbb{CP}^1$, the complex projective line.
For a more complete picture, think of how else you could define a Möbius transformation matching your requirements. You can define one which maps $x_0$ to $F(x_0)$, maps their inverse in the unit circle in the same way (which means you'll likely need homogeneous coordinates to deal with the point at infinity, the inverse of the center), and maps an arbitrary point on the unit circle onto itself (i.e. a fixed point). This choice still ensures that $x_0$ maps as requested, and preserves the unit circle. The choice of the point on the unit circle is arbitrary, so you have one real degree of freedom there to choose your transformation.
how to apply that to the points in Poincare disc model
Depending on your formalism this shouldn't be too hard. If you see the Poincaré disk as complex numbers $z\in\mathbb C$ with $\lvert z\rvert<1$, then this is a natural environment to apply a Möbius transformation $z\mapsto\frac{az+b}{cz+d}$. If you have points $(x,y)\in\mathbb R^2$, treating them as complex numbers $x+iy$ is the way to go.
Personally I have a background in projective geometry, so I tend to use homogeneous coordinates and write the Möbius transformation as $\scriptstyle\begin{pmatrix}z\\1\end{pmatrix}\mapsto\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}z\\1\end{pmatrix}$ but explaining homgeneous coordinates is probably better left to a different question.
without having points land outside of the disc
If you choose the transformation correctly, this should not be a problem, as a proper hyperbolic isometry will preserve the unit circle. So this is something to keep in mind when defining (or computing) the transformation, but not when applying it.
Sometimes it can be useful to model orientation-reversing isometries as Möbius transformations which preserve the unit circle but exchange the inside with the outside of the unit disk. To go back to a standard model, you'd apply an inversion in the unit circle afterwards. But since you asked about translations and those are orientation-preserving, this is a non-issue here.
What operation F do I apply to those points
In case the above helped you find a transformation, and you want to check your result: the Möbius transformation which satisfies the additional constraint mentioned above would be (mouse-over to show spoiler):
$$z\mapsto F(z)=\frac{2z-(1+i)}{-(1-i)z+2}=\frac{(1+i)z-i}{-z+(1+i)}$$
