Algebraic geometry is quite new for me, so this question may be too naive. therefore, I will also be happy to get answers explaining why this is a bad question.


I understand that the basic philosophy begins with considering an abstract commutative ring as a function space of a certain "geometric" object (the spectrum of the ring). I also understand that at least certain types of modules correspond to well known geometric constructions. For example, projective modules should be thought of as vector bundles over the spectrum (and that there are formal statements such as the Serre-Swan theorem which make this correspondence precise in certain categories). My question is, what is the *general* geometric counterpart of modules? 


This is not a formal mathematical question, and I am not looking for the formal scheme-theoretic concept (of sheaves of certain type and so on), but for the geometric picture that I should keep in mind when working with modules.


I will appreciate any kind of insight or even just a particularly Enlightening example. 



What is the general geometric interpretation of modules in algebraic geometry? 
Roughly a module can be thought of as a vector bundle on the spectrum, where the dimension of fibers may vary. Let me give some examples and facts:


* A free modules corresponds to trival vector bundles, or more generally projective modules correspond to vector bundles as you already pointed out.
* Let $R$ be the coordinate ring of a variety and $I$ a radical. Then the $R$ module $R/I$ corresponds to attaching a one dimensional vector space on each point of $Z(I)$ and the zero vector space everywhere else. For Example $R=k[x,y]$ and $I=(x,y)$ gives the skyscraper sheaf at the origin. $I=(x)$ gives the trivial one dimensional bundle on the y-axis etc. If your Ideal is not a radical, the situation is slightly more complicated. $R/I$ can be thought of as the trival bundle on an infinitesimal neighborhood of Z(I).
* Another nice example is a geometric explanation why the tensorproduct $\mathbb Z/p \otimes\_{\mathbb Z} \mathbb Z/q$ for say $p,q$ without common divisor vanishes. Our space
$spec(\mathbb Z)$ consists just of a point for each prime (and a generic point). Now with the above intuition in mind,
our two modules are geometrically just one dimensional vectorspaces attached 
to infinitesimal neighborhoods of the prime divisors of $p,q$. Since $p$ and $q$ have no common divisor there is no point where both $\mathbb Z/p$ and $\mathbb Z/q$ have nonzero fiber.
As with vectorbundles, tensor produduct of modules can be thought of geometrically as fiberwise tensor product ($i^\*$ commutes with $\otimes$). But of course the fiberwise tensor product vanishes because there are no points where both modules have nonzero fibers, so $\mathbb Z/p \otimes \mathbb Z/q=0$ .
* Finally any finitely generated module (more generally a coherent sheaf on a noetherian scheme) is built up of vector bundles on subspaces in the following way: There exists a stratification of spec(R) such that the module pulled back to the strata is a vector bundle. This follows from Hartshorne Ex II.5.8.


