A ring expansion typically occurs to overcome the high strain present in smaller sized rings.
Now strain isn't one single thing but a combination of many different factors which are nicely summarized by this[1] paper.
Types of strain
... the total strain of a conformation is the sum
of:
Bond strain—stretching or compression of chemical bonds. This type of strain is rather severe and is not encountered very often in
organic compounds. To minimize bond strain, a molecule adopts
conformations that have other, less energy-demanding, types of strain.
Torsional strain (eclipsing strain, Pfitzer strain) is caused by eclipsing interactions. Torsional strain is considerably higher
compared to steric or angle strain, which are explained next.
Steric strain (Van der Waals strain, Prelog strain) is caused by atoms forced too close to each other. Transannular strain (Prelog
strain) is a form of steric strain characteristic of medium rings.
Angle strain (Bayer strain, classical strain) is a result of deviation from the ideal bond angle. Compared to other types of
strain, increase in energy of a conformation caused by angle strain is
relatively low. As a result, a molecule can accommodate relatively
large deviation from an ideal bond angle and still be stable.
Different rings undergo expansion to overcome different types of strains. The major driving forces in smaller rings is the angle strain and the eclipsing strain.
Small rings are rigid and highly strained. They are characterized both
by a large deviation from the ideal tetrahedral valency angle—a high
angle strain, and eclipsing interactions—a high torsional strain.
For example, cyclobutane expands to cyclopentane because of reduction in angle stain and torsional strain.
But in case of cyclopentane's expansion into cyclohexane, the expansion occurs mainly due to reduction in torsional strain as angle strain actually increases in case of cyclohexane.
Angles in a regular pentagon are 108°. That is very close to the ideal
tetrahedral angle of 109.5° and, therefore, if cyclopentane were
planar it would not exhibit any significant angle strain. However, in
a planar cyclopentane there would be a high degree of torsional
strain. All ten of the carbon–hydrogen bonds would be in eclipsing
conformations.
Regarding the mechanism, the following reaction pathway is followed to obtain the product:
Later OP proposed a mechanism of their own and asked in the comments to verify it:
The mechanism you propose looks a bit incorrect. The arrows are pointing in the opposite direction. I have never seen an arrow hanging the $\ce{H+}$ to an $\ce{-OH}$ used to signify protonation.
I believe you are confused in the process of identification of nucleophile and electrophile. The arrows are pointed from the nucleophile to the electrophile. The arrows signify the movement of electrons. You might benifit from reading Drawing Organic Mechanisms.
Reference:
(1) Dragojlovic, V. Conformational Analysis of Cycloalkanes. ChemTexts 2015, 1 (3). https://doi.org/10.1007/s40828-015-0014-0.