The question you are asking, generally, does not have a trivial answer. It is not sufficient to just take a differential form on the plane and then "lift" it up to the surface in question, as, generally, you will not obtain a well-defined form in this way. However, it can be done in special circumstances. For example, if your surface is a quotient of the plane by some Fuchsian group of transformations, for example a group of translations, then you can take any differential form on the plane which is invariant under the given Fuchsian group, and lift it up to your surface. For example, a doubly-periodic form on the plane can be lifted to a form on the torus. In general, for more complicated surfaces and non-constant forms, you cannot do this.
Elliptic integrals yield examples of non-closed complex 1-forms on the torus. To understand why this is true, and what the torus has to do with elliptic integrals, you would need some basic understanding of the theory of elliptic functions. But you can imagine taking a constant vector field in the plane (which is essentially a 1-form, up to a choice of basis for the cotangent bundle), and lifting it up to the torus. You get a "hairy" torus, and the differential form which corresponds to this vector field under some choice of basis for the cotangent bundle of the torus is a non-closed form on the torus (imagine integrating the form around a circle going through or around the hole of the torus; you would not get 0 for both of these paths).
For a 1-form to be exact simply means that it is the gradient of a scalar field. For example, if you are given a surface with temperature, the gradient of the temperature is an exact form. The general definition is similar.