We want $U^{-1}M$, the localization of $M$ at $U$, to consist of "fractions of $M$ over $U$"; a good start toward this definition is to consider $M\times U$, writing $m/u$ for $(m,u)$. However, this is not enough: as with ordinary fractions, we expect there to be multiple ways of writing the same fraction ($1/2 = 2/4$). Thus we have to impose some equivalence relation. Let's try to reverse-engineer such an equivalence relation. Begin with two fractions that are equal: \frac mu = \frac{m'}{u'} and "cross-multiply" to get mu' = mu. This is a relation that can be checked in the original module $M$, and we might naively try to use this as an equivalence relation. However, we have forgotten that elements of $U$ can be cancelled; i.e. for $v\in U$, we have \frac mu = \frac{m'}{u'} = \frac{vm'}{vu'} giving the equation mvu' = vm'u, or v(mu' - m'u)=0. Thus we must weaken our equivalence relation on $M\times U$ to (m,u)\sim(m',u')\iff v(m'u-mu')=0 for some $v\in U$. It turns out that this is sufficient to make $M\times U/\sim$ a module in the "right" way. (satisfies some universal property, is equal to $M\otimes_R U^{-1}R$, etc.)
Remark: The "naive" definition suffices when $vn=0$ implies $v=0$ or $n=0$ for $v\in U$, $n\in M$; for example, when $M=R$ and $R$ is an integral domain. In general, though, I'm not sure that the "naive" definition even gives an equivalence relation (I'll check later, but I'm pretty sure that it doesn't).