How can I add 2 squares geometrically to get a bigger square? Suppose all I have is 2 square pieces of paper of equal size and a pair of scissors. How can I cut the paper and rearrange the pieces into 1 bigger square (combined size)?

I presume it would include some folding if given that the only unit of measure available is the paper itself.
The answer simply is to overlap the squares evenly and make one corner-to-corner cut. You'll get 4 right triangles of equal size whose hypotenuses can be rearranged as the sides of your new square.  Thanks to André and Thomas for pointing this out.

Alternatively, you can cut one square into 4 even right triangles (using 2 diagonal cuts), and place their hypotenuses against the sides of the remaining square. (Thanks Rahul)

The Wallace-Bolyai-Gerwien Theorem states that given any 2 polygons of equal area, either one can be cut into a finite number of smaller triangles and rearranged into the shape of the other polygon.  Since we can lay the squares side-by-side as a 1 by 2 rectangle, then by extension the theorem allows us to get to a square of equal size.
The minimum is to cut both squares each in 8 pieces, and make a square of 4x4. To do that fold the square four times (in 8 pieces) and then cut it.

If use both squares without cutting them you can't make one square because you would get a rectangle of 2x1. Same story in case you cut each in half: you would get 4 rectangles and no way to connect the four of them to make a square. If you cut each in 4 smaller squares you would get 8 squares total and since the area of a square is L x L, or in this case 8^(1/2)*8^(1/2), cutting both in 4 parts is not the solution: you can't cut a square paper in 8^(1/2) parts. So the minimum is to cut each square paper in 8 pieces (smaller squares): you would get 16 squares which is basically a 4x4 square.

PS: cutting the diagonal of each square will aldo work but you would get a hole of a smaller square in the center of the big square.