There is an intuitive explanation for that.
Warm up
We are trying to integrate the area of a sphere with radius r in spherical coordinates. The angle $\theta$ runs from the North pole to South pole in radians. Angle $\theta$ equals zero at North pole and $\pi$ at South pole. The distance on the surface of our sphere between North to South poles is $r \, \pi$ (half the circumference of a circle).
Lines on a sphere that connect the North and the South poles I will call longitudes.
In order to calculate the area of a sphere we cover its surface with small RECTANGLES and sum up their total area.
Explanation
In baby physics books one encounters this expression. In each infinitesimal rectangle the longitude component is its vertical side. The latitude component is its horizontal side.
$$\int_{0}^{ \pi }\int_{0}^{2 \pi } r^2 \sin {\theta} \, d\phi \,d\theta = \int_{0}^{ \pi }\int_{0}^{2 \pi }
\overbrace{
\underbrace {r \, d\theta}_{\text{longitude component}} *\underbrace {r \, \color{blue}{\sin{\theta}} \,d \phi}_{\text{latitude component}}}^{\text{area of an infinitesimal rectangle}}
$$
We see that the latitude component has the $\color{blue}{\sin{\theta}}$ adjustment to it. This is key. Why is that?
It is because rectangles that we integrate look like ordinary rectangles only at equator! Where $\color{blue}{\sin{\frac{\pi}{2}} = 1}$, i.e. where we do not need to adjust the latitude component. Near the North and South poles the rectangles are warped. Here is the picture.
![enter image description here](https://cdn.statically.io/img/i.sstatic.net/8aeGZ.png)
The blue vertical line is longitude 0. The brown line on the right is the next longitude to the east. Blue triangles, one at each pole and two at the equator, have markings on them. These markings represent equal angles for $\theta \, \text{and} \, \phi$.
We need to shrink the width (latitude component) of integration rectangles that lay away from the equator. Because only at equator they are not distorted. Why we choose the sine function? Intuitively, because its value goes from zero to 1, and then back to zero. That is, where $\theta$ and radius $r$ map out the zero longitude (part of a circle of a plane).
What happens when we drop this sine adjustment for the latitude?
Case B: drop the sine adjustment for the latitude
In this case all integration rectangles will be regular undistorted rectangles. Their total length along a longitude will be $r \, \pi$ and total length along the equator latitude will be $r \, 2\pi$. Total area will be $$r \, \pi \times r \, 2\pi = 2 \pi^2 \, r^2$$
Like this
$$\int_{0}^{ \pi }\int_{0}^{2 \pi } r \, d\theta * r \, d \phi = 2 \pi^2 r^2$$
Here is the picture.
![enter image description here](https://cdn.statically.io/img/i.sstatic.net/g4dWl.png)