why does ring flip of L-fructose occur?
These ring-flips are called 1C4-4C1 transitions and these happen to all of the pyranose sugars in solution when the conditions allow. However, the abundance of each forms depends on their energy. There is a so called 1,3-diaxial interaction between the axial substituents of a cyclohexane analogue. Two axial substituents on the same side are much closer than two equatorial substituens. If these substituents are relatively big, it will be energetically unfavorable to bring them close together due to steric effects. Therefore the molecule will "choose" the conformation that has less energy, so less "large" substituents on the same side. In your case the OH and Me groups are considered the "large" substituents, while the H atoms (undrawn) are the "small" ones. Therefore the conformation you drawn has an unfavorable Me-OH interaction, which will drive the conformation change from 4C1 to 1C4.