If the Moon's orbit were fixed, the area covered by the Moon would just be $360^{\circ}*(0.5^{\circ}+2*1^{\circ})$. The 5.3 degree inclination relative to the ecliptic (or some other inclination relative to the equator) would not be a factor.
What is important to this question about the coverage is that the orbit changes over time. In particular, the position of the ascending node (where the Moon's orbit crosses the ecliptic) moves along the ecliptic with a period of 18.6 years (Wikipedia article on Lunar precession).
In a plot of the sky using ecliptic coordinates (so that the ecliptic is a straight line), the 5.3 degree inclination of the Moon's orbit to the ecliptic is shown as "year X" in my figure. The precession changes the orbit to positions such as year Y and year Z in my figure, and during the entire 18.6 year cycle, the entire band within 5.3 degrees of the ecliptic is covered by the Moon. In other words, the ascending node moves through all 360 degrees of ecliptic longitude. This results in the term 2*5.3 in your equation. (Note: I do not know if the precession is to the "right" or "left", so do not interpret my figure to mean that year Y is later than year X. Year Y could be earlier than X. I'm sure someone will correct me!)

If the same plot were made in equatorial coordinates (so that the equator is a straight line), the band covered by the Moon is of course the same on the sky. However, the precession along the ecliptic limits the ascending node of the Moon's orbit on the equator to a small range or right ascension; the ascending node does not travel around the entire equator. Representing that band using equatorial coordinates and determining the area inside the bands is much more difficult.
