Without looking at the clues or discussion. This isn't optimized, but it works (I think):

There are only two hand placement positions, adjacent and opposite.

1. Allow a round to pass without touching anything, to confirm that they don't all start the same.

2. Place hands adjacent.

If you get two of the same orientation,

then flip both. Now you either win, or there are three of the same orientation remaining

else (you get two with different orientations)

flip them so they're the same. Either you win, or there's at least one of the other orientation.

3. Place hands opposite. One of the bowls you get is guaranteed to be one from the previous step. If the other is in the same orientation, then flip nothing. You have three bowls in the same orientation. If the other is in the opposite orientation, flip it so now it's the same. Either you win or you have three bowls in the same orientation.

3/4. Place hands adjacent again. Either you get the single remaining in the opposite orientation, flip it and win. Otherwise, you get two in the same orientation. Flip 1.

4/5. Now there are two pairs in the same orientation. Place hands opposite. If you get two of the same orientation, flip both and win, otherwise there are two pairs of adjacent bowls with the same orientation. Flip nothing.

5/6. Place hands adjacent. If you get two of the same orientation, flip both and win. Otherwise, flip both.

6/7. Now there are two pairs of opposite bowls with the same orientation. Place hands opposite. Flip both and win.