Somewhat more general answer, since the specifics are well taken care of. Warning: Math ahead.
Label each of the positions around the clock from 1 to 12, so that in the 'correct' clock 1 is at position 1, 2 is in position 2 and so on. Label the center of the clock '0'.
Note that any sequence of moves which could possibly put the clock into the correct position (assuming the center of the clock starts out empty) can be divided into a number of segments, each of which takes the form:
1. Move number at 3, 6, 9, or 12 to center.
2. Rotate every number around the circle exactly once either clockwise or counterclockwise, leaving the same space at 3, 6, 9, or 12. Repeat zero to five times.
3. Rotate a group of three, six, nine, or twelve numbers adjacent to the position you just moved the one space either clockwise or counterclockwise.
4. Return center number to the gap, which will have shifted three, six, nine or twelve spaces around the board, therefore still being reachable from the center.
Note that, if we write out the list of numbers from position 1 to position 12, for example (12, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 1) in the question's starting case, moves of this kind can do one of three things:
If we return the center number to the same gap, then we have just cycled the rest of the list around. For example, in the question's starting position, if we move the 1 into the center, shift the other numbers clockwise twice and put the one back at position 12, we get:
(10, 11, 12, 2, 3, 4, 5, 6, 7, 8, 9, 1).
If we return the center number to a different gap without rotating every number, then we have shifted the numbers between the two gaps (direction depending on the direction we rotate). For example, putting one into the center, shifting 12, 2, and 3 counterclockwise and moving 1 to the newly opened position 3 gives us:
(2, 3, 1, 4, 5, 6, 7, 8, 9, 10, 11, 12).
If we do both, we combine the two effects, shifting all the numbers and then just the ones in between. For example, if we move one to the center, shift everything clockwise four times, then put the gap at position 6 by shifting the numbers at 6, 5, 4, 3, 2, and 1 counterclockwise once, we go from:
(12, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 1)
(8, 9, 10, 11, 12, 2, 3, 4, 5, 6, 7, 1)
(9, 10, 11, 12, 2, 1, 3, 4, 5, 6, 7, 8)
Note that this is just a combination of the other two types.
Now let's do a bit of math- specifically, let's look at Permutations.
We can think of our lists as being a permutation written in cycle notation, and our sequences of moves as being more permutations applied to the existing one. As a couple examples:
The moves we make in the first example of the above section can be written in cycle notation as the permutation:
(10, 1, 12)
The moves we make in the second example of the above section can be written as the permutation:
(12, 1, 4)
The moves in the third example can be written as a combination of:
(8, 1, 12)
(1, 8, 3),
which is just:
(12, 3, 1)
More generally (proof not hard, but left to reader for brevity), we can say that:
Moves of the first kind where we shift $n$ numbers clockwise result in permutations of the form $(a, b, c)$ where $a$ is the number moved to the center, $b$ is the number that was immediately clockwise to it initially, and $c$ is the number that was $n$ spaces counterclockwise to it initially (and ends up immediately clockwise to it after the move). Counterclockwise is similar.
Moves of the second kind where we move the number $a$ at position $m$ to position $n$, rotating numbers that are clockwise from position $n$, where $n, m$ are 3, 6, 9, or 12 and $n \neq m$, give permutations of the form $(a, b, c)$ where $b$ is the number at $m$ and $c$ is the number immediately clockwise from $n$. Counterclockwise is similar.
Note that a property of permutations is that they can be decomposed into transpositions. In our case, this decomposition is especially simple:
$(a, b, c)$ becomes $(a, b)(a, c)$
A word of caution- since permutations have no set starting position, and for example (1, 2, 3) (2, 3, 1) and (3, 1, 2) are all the same, we need to be cautious in finding solutions- it is not sufficient that the end result of the permutations is (1, 2, 3, 4... 10, 11, 12), it must also be the case that the number 12 is at position 12, etc. To ensure this is possible, given our moves defined above (A rigorous proof is again not hard, but omitted for the sake of brevity):
Even if $(a, b, c)$ does not correspond to some valid move of the first type, note that for any $d$ it is true that $(a,b,d)(a,d,c) = (a,b,c)$, and for any $e,f,g$ it is true that $(e,f,g)(a,b,d)(e,g,f)(a,d,c) = (a,b,c)$. Additionally, if $(e,f,g)$ is a valid move of any kind, or if there is an equivalent set of moves which ends by moving $e$ to position 3, 6, 9 or 12, then once $(e,f,g)(a,b,d)$ is applied $(e,g,f)$ is guaranteed to be valid as long as none of $a,b,d$ are equal to $e,f,g$.
In particular for the given starting set (proving Angkor's result, because why not):
Applying (1,12,2) (a type 1 move) would instantly fix the order, as noted by Sechiro. This is equivalent to (12,2,1), which would only work if 12 was already at the top with a 2 next to it, which is equivalent to (12,2,9)(12,9,1), which is the same as (3,4,11)(12,2,9)(3,11,4)(12,9,1) if (3,4,11) is a valid type-2 move since it would ensure the conditions of (12,2,9) are met. It isn't valid since 11 isn't at 3, 6, 9 or 12, but (3,4,2)(3,2,11) at long last is valid, and is equivalent to it, so the full sequence (3,4,2)(3,2,11)(3,4,11)(12,2,9)(3,11,4)(12,9,1) keeps the order of (1,12,2), while ensuring 12 ends up at the top where it belongs.
This method could stand some fine tuning as it still relies a bit on intuition and while it happens to come across an optimal solution in this case I can't think of a reason why it should in general, but it generalizes across any starting or ending position and it should generalize easily with some proof finagling across any layout of clock.