Since F is a K module and K is a division ring, F is always a vector space over K. The dimension of the extension is the dimension of the vector space. The extension is finite if the resulting vector space has finite dimension. A quadratic extension has dimension 2.
The product of the dimensions of two consecutive field extensions is the dimension of the composite extension. This follows directly from the corresponding theorem for modules over a division ring.
Let F be an integral domain that contains the field K. Since F is a K module, it is a vector space over K. If the dimension of F is infinite, F need not be a field. Let K be the reals and let F be K[x], polynomials with real coefficients. The powers of x can act as a basis for this vector space. This is an integral domain that is not a field.
However, if F/K is finite, it is a field. Let x be a nonzero element in F and consider the map x*F. By cancelation, the map is 1-1. It is also a K module homomorphism. In other words, the map preserves the essence of addition and scaling by elements in K. Since we are dealing with vector spaces, the map is a linear transformation. With an empty kernel, the dimensions of the domain and range are equal. Our 1-1 map is onto, and x has an inverse in F. This holds for all x, hence F is a field.
This generalizes to a (possibly infinite) algebraic extension. Since K(x) is a finite extension for every x, the above proof holds. Every x is invertible and F is a field.