Why and how does ring expansion occur in the dehydration of (cyclobut-3-ene-1,2-diyl)dimethanol?

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:

1. 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.

2. 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.

3. 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.

4. 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.}

The mechanism for the given reaction would be as follows.

Why does this happen? This is because a four member ring has a high amount of ring strain. Expanding itself to a 5 member or (if possible) a six member ring is highly advantageous as it helps alleviate some of this ring strain.

Now, in this case there is also the added advantage of the six member ring being aromatic (benzene).

A similar example has been discussed here: Formation of toluene from (cyclobutane-1,2,3-triyl)trimethanol using conc. sulphuric acid