Define the run-partition number of a permutation $\pi$, denoted $r(\pi)$, using the following process. Let $k$ be the maximal integer such that the numbers $\min(\pi),\ldots,k$ appear in increasing order in $\pi$. Remove them from $\pi$, and repeat the process. The number of rounds it takes to consume the entire permutation is $r(\pi)$.
For example, let's compute $r(62735814)$. We first set aside $1$, to get $6273584$. Then we set aside $234$, to get $6758$. Then we set aside $5$, to get $678$. Finally, we set aside $678$ to get the empty permutation. This takes four rounds, so $r(62735814) = 4$.
How is this representation useful for sorting $62735814$? Take every second run, i.e. $234,678$, and move these numbers to the right to get $51627384$ (edit: in the order they appear in the permutation, i.e. $627384$). Now there are only two runs, namely $1234,5678$, and we can sort the list by moving $5678$ to the right.
Now let me make the following conjecture: For a permutation $\pi$, let $\Pi$ be the set of all permutations that are reachable from $\pi$ within one move. Then $\min_{\alpha \in \Pi} r(\alpha) = \lceil r(\pi)/2 \rceil$.
Given this conjecture, it is easy to show that the minimal number of moves needed to sort a permutation $\pi$ is $\lceil \log_2 r(\pi) \rceil$, and I have verified this formula for all permutations in $S_n$ for $n \leq 8$.
Edit: Here is a different interpretation of the run-partition number which gives a linear time algorithm for computing it, and allows me to sketch a proof of my conjecture, thus verifying the formula $\lceil \log_2 r(\pi) \rceil$.
Consider the permutation $62735814$ again. The reason that the first run ends in $1$ is that $2$ appears before $1$. Similarly, the second run ends in $4$ since $5$ appears before $4$, and so on. Therefore the run-partition number of a permutation is the number of $i$s such that $i+1$ appears before $i$.
We can state this more succinctly if we look at the inverse of the permutation. Consider again $\pi = 62735814$. Take $\pi^{-1} = 72485136$. This permutation has three descents: $7\mathbf{2}48\mathbf{5}\mathbf{1}36$ (a descent is a position smaller than the preceding one). Each of the descents corresponds to the start of a new run. So $r(\pi)$ is equal to one plus the number of descents in $\pi^{-1}$.
What does the operation look like in terms of $\pi^{-1}$? let $B$ be the set of numbers that we move to the right, and $A$ be the set of numbers that stay to the left. We replace the numbers in $A$ with a permutation on $\{1,\ldots,|A|\}$ representing their relative order, and replace the numbers in $B$ with a permutation on $\{|A|+1,\ldots,|A|+|B|\}$. For example, consider the move $\mathbf{6273}5\mathbf{8}1\mathbf{4} \mapsto 51\mathbf{627384}$. In terms of the inverse permutations, it's $7\mathbf{248}5\mathbf{136} \mapsto 2\mathbf{468}1\mathbf{357}$. So $75$ was mapped to $21$ and $248136$ was mapped to $468357$.
A descent $\ldots xy \ldots$ in $\pi^{-1}$ is lost after the operation only if $x \in A$ and $y \in B$. Conversely, in terms of $\pi^{-1}$, the partition into $A$ and $B$ corresponds to $A$-runs and $B$-runs; every time a $B$-run ends and an $A$-run begins, there is a descent. In order to "kill" a descent, we have to switch from an $A$-run to a $B$-run. If we kill two descents, we will have switched in the middle from a $B$-run to an $A$-run, incurring a descent.
This argument can be formalized to show that if $\alpha$ arises from $\pi$ via an operation, then $d(\alpha^{-1}) \geq \lfloor d(\pi^{-1})/2 \rfloor$, where $d(\cdot)$ is the number of descents. This is equivalent to $r(\alpha) \geq \lceil r(\pi)/2 \rceil$, thus proving one direction of my conjecture. The other direction is easier, and was already outlined above: we simply take every second run and push these runs to the right to get a permutation $\alpha$ satisfying $r(\alpha) = \lceil r(\pi/2) \rceil$.
