That is a pretty terse proof! Let me give an outline of a proof that I know. First, one could deduce the statement from a more general:

**Theorem 1**: Let $R$ be a regular local ring, $E$ be a reflexive $R$-module locally free on $U_R$, the punctured spectrum such that $E$ has no free direct summand. Then one can find a free module $T$ and a filtration:

$$E\oplus T = F_0 \supseteq F_1 \supseteq \cdots F_N =0$$

with $F_i/F_{i+1}$ a syzygy of $k=R/m$.

Why is this local statement implies what you want?

Let $A=k[x_0,\cdots, x_n],m=(x_0,\cdots,x_n), X=Proj(A)=\mathbb P^n, R=A_m$. There is natural functor from the category of vector bundles on $X$ to that of vector bundles on $U_R$, which is the same as the category of reflexive $R$-modules which are locally free on $U_R$. This is used by Horrocks all the time and is explained in Section 9 of his paper: "Vector bundles on punctured spectrum of a regular local ring".

A proof of Theorem 1 can be found in Chapter 5 (theorem 5.2) of the book "Syzygy" by Evans-Griffith. A brief outline in case you can't find the book:

As suggested in the paper you quoted, one starts with a minimal resolution of $E^*$. Then dualizing gives a complex (remember that $E^{**} \cong E$ as $E$ is reflexive):

$0 \to E \to L_0 \to L_1 \cdots $

whose cohomologies are $Ext^i(E^*,R)$. Let $i>0$ be the smallest number such that $X=Ext^i(E^*,R) \neq 0$ Break the l.e.s in to the exact sequences:

$0\to E \to L_0 \to L_1 \cdots \to L_i \to N \to 0 (*)$

and $0 \to X \to N \to N/X \to 0$. Now build free resolutions for $X$ and $N/X$ and map them onto $(*)$ as in Horseshoe Lemma, stopping at the spot $E$, one gets a s.e.s:

$0 \to B \to E\oplus T \to C \to 0$
here $T$ is free and $C$ is a syzygy of $X = Ext^i(E^*,R)$. Repeat if necessary and you have a filtration whose quotient are syzygies of various $ Ext^i(E^*,R)$. But each of this $Ext$ modules has finite length (as $E$ is locally free on $U_R$), so they can be filtered by copies of $k$. Now use the same trick to build a finer filtration whose quotients are syzygies of $k$. Since $R$ is regular, the resolution of $k$ is the Koszul complex, answering your second question.