Extending Banach-Tarski paradox? I've learned the Banach-Tarski paradox as following:


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*The points on the sphere (but not the fixed points) are drawn as a square grid, form each point there are three new directions plus the direction back to the point it came form. 

*The grid is cut in four pieces:

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*(A) The x-axis, plus the points where you can come when turning right

*(B) The points  where you can come when turning left, excluding x-axis

*The points where you can come by going up

*The points where you can come by going down



Then there is a new sphere created by moving all points on A one place to the left. This creates together with B the new sphere.
Now my question is: Can we create a complete sphere by moving all points on A countably infinitely many points to the left?
 A: The question is a bit light on details, but if I understand it correctly the answer is no.
First let me fill in some of the details of the usual argument.  We take a copy of $F_2$, the free group on two generators, contained in $\text{SO}(3)$, the group of isometries of the sphere $S^2$.  Excluding the set $D$ of fixed points (meaning points that are fixed by some nonidentity element of $F_2$) the action of $F_2$ on the remaining points is free.  There are only countably many fixed points, so there is a trick for dealing with these later; let's ignore them.
In $S^2 \setminus D$, we use $\mathsf{AC}$ to choose, for each orbit, a point in that orbit.  This allows us to consider each orbit of $S^2 \setminus D$ as a copy of $F_2$ and to transfer a paradoxical decomposition of $F_2$ to each copy simultaneously, giving a paradoxical decomposition of $S^2 \setminus D$.
The question seems to be considering a single orbit (copy of $F_2$) on its own, so it essentially asking about paradoxical decompositions of $F_2$.  (By the way, I think it is a bit misleading to say "square grid": in a square grid if I go right, up, left, and then down, then I am back where I started, whereas in $F_2$ if I multiply $a$, $b$, $a^{-1}$, and $b^{-1}$ in that order, then I do not get the identity element.)
Now for your question:

Can we create a complete sphere by moving all points on A countably infinitely many points to the left?

I don't think it makes sense to move "infinitely many points to the left".  In $F_2$ there is no element $b^{-1}b^{-1}b^{-1}\cdots.$  You might think that because we are consider a copy of $F_2$ that is contained in the sphere, there is a metric and so we could try to take a limit:
let $x \in S^2 \setminus D$ be the point we have chosen from this orbit and consider the sequence of points $x, b^{-1}x, b^{-1}b^{-1}x,\ldots.$  However, this sequence is not convergent (in fact its set of limit points forms a circle.)
You may be confused by diagrams that show $F_2$ contained in the plane, and in which points approach a limit as they move to the left:

However, the metric that $F_2$ inherits from the plane $\mathbb{R}^2$ in this picture is not the same one that it inherits from $S^2$.  In fact, this picture does not represent an isometric action of $F_2$ at all (note that the edges do not all have the same length.)  So the usefulness of this picture for understanding the paradoxical decomposition of the sphere is limited.
