Is there a functionality for finding how much two objects intersect each other

Well, if you constraint your problem so you always have two identical spheres it's relatively easy.

Two identical spheres, as long as they intersect, have a single "cutting plane" which is located exactly halfway between the two spheres. Just imagine you cut both spheres at this plane. You will get 4 sphere pieces. Each sphere breaks up into a larger (`>50%`) and a smaller (`< 50%`) piece. You will ignore the larger ones and simply look at the smaller ones. The smaller pieces are Sphere caps so you can simply calculate the volume of each piece and add them together to get the whole intersection volume. Since the spheres are identical, both caps are also identical. So you can just calculate the volume of one cap and multiply it by 2. The result can be divided by the volume of a whole sphere to get a value between 0 and 1.

Let "a" and "b" be the position of sphere A and sphere B. "r" is the radius of both spheres

 float IntersectSphereVolume(Vector3 a, Vector3 b, float r)
 {
     Vector3 ab = b - a;
     float halfLength = ab.magnitude * 0.5f;
     if (halfLength > r)
         return 0; // no intersection
     float circleRadiusSqr = r * r - halfLength * halfLength;
     float height = (r - halfLength);
     float vCap = (3*circleRadiusSqr + height*height) * height * Mathf.PI / 6f;
     
     float vIntersect = vCap * 2f;
     float vSphere = r*r*r*Mathf.PI * 4f / 3f
     return vIntersect / vSphere;
 }

Just written from scratch so i can't confirm it works properly ^^. This method returns a value betwen 0 and 1 where 0 means no intersection and 1 when the two spheres are at the exact same position.

If you don't need the exact volume matching you could simply use a linear error like this:

 float IntersectSphereVolume(Vector3 a, Vector3 b, float r)
 {
     Vector3 ab = b - a;
     return Mathf.Clamp01(1f - ab.magnitude / (2*r));
 }

This will also return a value between 0 and 1 but the curve will be linear.

ps: In the first method one could cancel out Mathf.PI and some other factors as we finally just divide both volumes through each other, However i kept it so it's clear what happens. Also you can use the volume for other things.

If the spheres have different radii or aren't spheres at all this is getting much more complicated and doesn't have a general solution.

This is an old question, but for concave meshes you need boolean operations, which is quite complex math - but luckly well understood.

If one of your meshes is concave and the other is convex, have a look at the Sutherland-Hodgman algorithm. However, this algorithm requeries one mesh to be convex. If you want both to be convace and also want to take care of edge cases, you need to go for boolean operations, such as the Martinez algorithm. The paper is called "A new algorithm for computing Boolean operations on polygons".