The problem:
Assume a bucket $B_{(1,0)}$ containing $n$ identical balls. If it is recursively split for $r$ rounds where $r \to \{0,1...\}$ so that at round $r$, we have $2^r$ buckets $B_{(1,r)}...B_{(2^r,r)}$ were $|B_{(i,r-1)}| = |B_{(i,r)}| + |B_{(2^{r-1}+i, r)}|$ and $ \frac{1}{3}|B_{(2^{r-1}+i,r)}| < |B_{(i,r)}| < \frac{2}{3} |B_{(2^{r-1}+i,r)}|$ for $1 \leq i \leq 2^{r-1}$.
(1) What is the minimum number of rounds needed to ensure exactly $m$ buckets with size $\le m$ where m is a predefined constant $1 \le m \le n$?
(2) What would be the maximum error be, if the error is define as $$ e := \sum_{i=1}^{2^r} | \lceil\frac{n}{m}\rceil - |B_{(i,r)}|| $$
(3) How may balls need to be removed before all m buckets are of equal size?
I've been giving this some thought, even though I haven't had time to do much. This is not a complete answer, but I'm hoping it will be useful.
Let $$p = \frac{|B_{(i,r)}|}{|B_{(i,r-1)}|}$$ Then $$1 - p = \frac{|B_{(2^{r-1}+i, r)}|}{|B_{(i,r-1)}|}$$ and when divided through by $|B_{(i,r-1)}|$, your inequality becomes $$\frac 13(1-p) < p < \frac 23(1-p)$$ $$1-p < 3p\quad\text{and}\quad 3p < 2 - 2p$$ $$\frac 14 < p < \frac 25$$ So your divisions are never even. The smaller bucket contains between 25% and 40% of the balls, while the larger bucket contains between 60% and 75%.
There are a number of issues that arise because of the discrete nature of this problem. For instance, what happens when a bucket gets down to just one ball? You can no longer divide the balls per the inequality, so the rounds must stop. A modification of the problem avoids these difficulties:
Instead of balls and buckets, start with a string of length $n$ laid out on a line. Instead of a round consisting of moving balls into new buckets, it consists of, starting with the leftmost string, cutting off an integral length of between 60% and 75% of the string and moving it to the rightmost end of the line. Continue until all existing strings at the beginning of the round have been cut (you don't cut the new bits until the next round).
This corresponds directly to your ball-and-buckets description, with each string being a bucket, and its length acting as the number of balls. But if we drop the "of integral length" requirement, we can always divide the strings to meet the inequality, no matter what their length is. So there is nothing to stop the rounds other than meeting or exceeding the requirement of exactly $m$ strings of length $\le m$. We can also change our unit of measurement so that the original string has total length $1$, and the requirement is now $m$ strings of length $\le \frac mn$. But as this is now the only use of the arbitrary value $n$, we can just replace $\frac mn$ with some $\lambda > 0$.
Whether the added issues of the discrete problem are something you are really interested in is your decision, not mine. But I am looking at the continuous version where I don't have to worry about them. You can consider it as just an approximation to the balls-and-buckets problem. I'd also like to clean up the symbolry a bit. I find that $B$ notation cumbersome. So:
The index $i$ of a string in round $r$ can be expressed as a binary number with $r$ bits (including leading zeros). By removing the highest order bit, one gets the index of the string in round $r-1$ from which the round $r$ string was cut. If the removed bit was a $0$, then the string is the shorter of the two pieces, while a removed bit of $1$ means it was the longer piece. Thus $b(i)$ is the number of times string $i$ descended from the larger piece in a cut.
The length can be expressed as $$L_r^i = \prod_{s=1}^r \rho_s^{i\bmod 2^s}$$ For each round $s$ where the string is descended from a larger piece, $0.6 < \rho_s^{i\bmod 2^s} < 0.75$, while for each round $s$ where the string is descended from a smaller piece, $0.25 < \rho_s^{i\bmod 2^s} < 0.4$.
Thus we can be sure that if $b(i) > b(j)$ then $L_r^i > L_r^j$. And in fact, $$\left(\frac14\right)^{r - b(i)}\left(\frac 35\right)^{b(i)} \le L_r^i \le \left(\frac25\right)^{r - b(i)}\left(\frac 34\right)^{b(i)}$$Next, let's examine the case where $m = 2^{r_0}$ for some $r_0$. On round $r$, the shortest possible longest string is $(3/5)^r$, while the longest possible string is $(3/4)^r$. So
On the next round, we would need exactly half the strings $\le \lambda$. A simple combinatorial argument shows that there are ${r \choose k}$ binary numbers $0\le i < 2^r$ with $b(i) = k$. When $r_0$ is even, that means you need all strings with $b(i) = \frac{r_0}2$ to be $\le \lambda$, while all strings with $b(i) = 1+\frac{r_0}2$ should all be $> \lambda$. Because $(3/5)^{r_0} \le \lambda$ (since we couldn't win in the previous round), the first condition is easily met. The maximum length for strings with $b(i) = 1 +\frac {r_0}2$ is $$\left(\frac25\right)^{r_0/2}\left(\frac 34\right)^{r_0/2 + 1}$$
When $r_0$ is odd, we need half of the $r_0 + 1\choose (r_0+1)/2$ strings with $b(i) = (r_0+1)/2$ to be $\le \lambda$ while the other half are $>\lambda$. Again, since $\lambda \ge (3/5)^{r_0}$, getting the needed strings $\le \lambda$ is easy. The maximum length for these strings is now $$\left(\frac 3{10}\right)^{(r_0+1)/2}$$ - if $(3/5)^{2r-1} \le \lambda < (3/10)^r$, where $m = 2^{2r-1}$, the game can be won in $2r$ rounds.
This can be generalized to $m$ that are not powers of $2$ and more rounds. To find out if the condition can be met on round $r$, assuming it couldn't be met on $r-1$, find the value $k$ such that $\sum_{i=0}^{k-1} {r\choose i} < m \le \sum_{i=0}^k {r\choose i}$. The game can be won on round $r$ if $$\lambda < \left(\frac25\right)^{r - b(k)}\left(\frac 34\right)^{b(k)}$$