Confuse on using bucket number to replace hashing multiple times in hyperloglog explanation
Let us start with the original approach. We compute the hash values of all input elements, and choose the minimal one $h$. If the leading bit is at position $k$, then we estimate that the number of distinct inputs in the output is roughly $2^k$.

The idea behind this is that we think of the minimal hash as a minimum of $N$ random strings, where $N$ is the number of distinct inputs. We expect the leading bit of such a minimum to be at position roughly $\log_2 N$.

When using $D$ different hash functions, the corresponding experiment is


  Choose $N$ random strings for each of $D$ many bins.


We then calculate the leading bit of the minimal string in each bin, and use this to estimate $N$.

Now suppose that we use the top $B$ bits to choose one of $2^B$ buckets, and compute the minimum value in each bucket. As before, this should be similar to the following experiment:


  Distribute $N$ random strings into $2^B$ bins at random.


As before, we then calculate leading bits and use them to estimate $N$, but since this step is identical to the corresponding step in the multiple hash method, we just concentrate on the first part of the experiment (putting random strings into bins).

We expect each bin to get roughly $N/2^B$ random strings, and so this experiment is very similar to the following one:


  Distribute $N/2^B$ random strings for each of $2^B$ many bins.


This is the same as the experiment for multiple hash functions, with $D = 2^B$ and $N$ reduced by a factor of $2^B$.

You might be put off by the difference in $N$. However, this difference disappears when we calculate the number of hash function evaluations. If we have a quota of $M$ hash function evaluations, then in the multiple hash case we can only use $M/D$ inputs, whereas in stochastic averaging we can use $M$ inputs.