How does one cut onions in a mathematically efficient way?

If we can rearrange the pieces any which way between cuts, as the question states, then we might as well arrange them in a line before each cut, so that the cut intersects every piece of the onion. By choosing each piece's translation perpendicular to the cut, we can guarantee that it is divided into two pieces of equal area. By choosing the piece's orientation, we can also ensure that none of the pieces get too long and skinny, so the diameter of a piece of area $a$ is bounded above by $c\sqrt a$, where $c$ is a universal constant.

Then it is sufficient to get the area of each piece below $M^2/c^2$. The original area of the largest piece is $4\pi r^2$, and each cut reduces it by half, so we can succeed in $\log_2(r^2/M^2)$ cuts plus a constant.

The converse argument shows that $\log_2(r^2/M^2)$ cuts plus a constant are indeed necessary.

With parallel cuts along planes spaced by 1/2cm in each of the three spatial dimensions. First, cut the peeled onion down the middle. Second, put one half on a flat surface and, from the (top) end opposite the root, make several cuts parallel to the surface most of the way into the onion (leaving a bit near the root to hold it together). Third, from the top make several parallel vertical cuts perpendicular to the previous cuts. Fourth, from the 'top' finish dicing the onion by making several parallel vertical cuts perpendicular to the previous two sets of cuts. Fifth, do the same with the other half.