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

• 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:
• (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?

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.)

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.