A "simple" approach is to memorize objects as volumes (for 2 spheres you just have 3 coordinates + radius), then you sample points until you find a "solid" point after that you start searching only nearby points where there's "empyt/solid" surface and you generate vertices in thhose points (you can link vertices using something like marching cubes).

note that I just proposed you the dummiest and not optimized algorithm, there are for sure much more better ways to do that.

note that there exist Framworks for doing that already (any 3d editing tool like Maya or blender already do that at application/script level), so if you find one for you language you can use it directly and probably will result in something more optimized than anything you can do youself

The short version is that there's no magic here -- You remember all those algebra questions from high school where the instructor asked you to find the point where two lines intersect? Its basically just like that, except you have more complex 3-Dimensional shapes involved. If you can describe the shape mathematically, and you have another mathematically-defined shape in the same vector space, then in principle you can plug in 2 inputs into both equations (say, X and Y), and see if the third comes out the same (Z), if so, that point is an intersection of the two shapes.

Of course, simply sampling points like this is inefficient, so they do things more cleverly, but the optimized forms are all based on this basic fact. For example, lets say you wanted to calculate the intersection of a cube and a sphere -- well, a sphere is about as simple as it gets, having a center and a radius, and a cube is just 6 planes that are constrained; there are well-known algorithms for finding out if and how a sphere intersects a plane: Take any plane in X-Y, and if the sphere intersects it, the center of the resulting circle is Xs,Ys,Zp (where sub-p means 'plane', and sub-s means 'sphere') -- Now, you can use the Pythagorean theorem the find the radius of the circle that's struck on the plane by the sphere -- The longest-side-term (hypotenuse of the right triangle) is given by the radius of the sphere, and one of the sides is given by the distance between the center of the sphere, and the center of the circle we calculated earlier. By the way, if the distance between the centers is longer than the radius of the sphere, you know the sphere doesn't interact with that plane. Now you have a description of exactly how the sphere interacts with that plane, and you can now apply the same constraints to the circle as the side of the cube (this is a similar process, except now you're operating inside a plane, and the former 'plane' is now a line), and if you still have some circle left, you have a description of how the sphere interacts with that side of the cube. Repeat for all 6 sides, and you're done. You know a lot about how the sphere and the cube interact and you can use that information to modify the shapes involved.