Here is a summary of what I learned from a nice expository account by Eisenbud (written in French), can be found as number 27 [here](http://www.msri.org/~de/papers/index.html). 

First, one studies a more general problem: Let $A$ be a Noetherian ring, $M$ a finite presented projective $A[T]$-module. When is $M$ *extended* from $A$, meaning there is $A$-module $N$ such that $M = A[T]\otimes_AN$?

The proof can be broken down to 2 punches:

**Theorem 1** (Horrocks) If $A$ is local and there is a monic $f \in A[T]$ such that $M_f$ is free over $A_f$, then $M$ is $A$-free 

(this statement is much more elementary then what was stated in Quillen's paper). 


**Theorem 2** (Quillen) If for each maximal ideal $m \subset A$, $M_m$ is extended from $A_m[T]$, then  $M$ is extended from $A$  

(on $A$, locally extended implies globally extended).

So the proof of Serre's conjecture goes as follows: Let $A=k[x_1,\cdots,x_{n-1}]$, $T=x_n$, $M$ projective over $A[T]$. Induction (invert all monic polynomials in $k[T]$ to reduce the dimension) + further localizing at maximal ideals of $A$ +  Theorem 1 show that $M$ is locally extended. Theorem 2 shows that $M$ is actually extended from $A$, so by induction must be free.

Eisenbud note also provides a very elementary proof of Horrocks's Theorem, basically using linear algebra, due to Swan and Lindel (Horrocks's original proof was quite a bit more geometric). 

As Lieven wrote, the key contribution by Quillen was Theorem 2: patching. Actually the proof is fairly natural, there is only one candidate for $N$, namely $N=M/TM$, so let $M'=A[T]\otimes_AN$ and build an isomorphism $M \to M'$ from the known isomorphism locally. 

It is hard to answer your question: what did Serre miss (-:? I don't know what he tried? Anyone knows?