A fixed presentation computes sections after every flat-family pullback

For a flat-family pullback put $K_B=\ker (E_{0,B}\to F_B)$ and $L_B=\ker (E_{1,B}\to K_B)$. Right exactness makes $E_{1,B}\to K_B$ onto. Both kernels are base-flat by the Tor argument, since $E_0,E_1,F_B$ are base-flat. They are finitely presented as follows. Write $B$ as the filtered colimit of its finitely generated $\mathbb{Z}$-subalgebras containing a finitely generated stage over which the presentation of $F$ descends (these stages are Noetherian). By the flat descent in [F2], after passing to one stage the given pullback $F$ is flat there. At that stage both successive kernels of the pulled-back presentation are coherent and base-flat. Tensoring these two sequences with $B$ stays exact because their quotients are base-flat. It identifies the stage kernels' pullbacks with $K_B,L_B$, so these kernels are finitely presented. Their fibre polynomials are respectively $P_{E_0}-P$ and $P_{E_1}-P_{E_0}+P$, independent of $B$.

By [F1], one bound depending only on these polynomials and $E_0,E_1$ makes every fibre of $K_B,L_B$ regular. Increase it to make $E_0,E_1$ regular too. The relative result gives $H^1(K_B(r))=H^1(L_B(r))=0$, so $H^0(F_B(r))$ is the cokernel of $H^0(E_{1,B}(r))\to H^0(E_{0,B}(r))$, for every $r$ beyond this bound. The two twist-section modules themselves commute with all base changes and are finite free in this degree range.

Over the original Noetherian base, put $K=\ker (E_0\to F)$ and $L=\ker (E_1\to K)$. Increase $N$ further so $H^1(K(r))=H^1(L(r))=0$ for $r\ge N$, by [F2]. The original $H^0(F(r))$ is now the same presentation cokernel. Right exactness of tensoring, together with the twist base-change isomorphisms, identifies its tensor with the cokernel in step 2.1. This is the canonical base-change map since all maps came from the given presentation. The relative result makes the target locally free of rank $P(r)$. Thus $N$ is independent of the base-change algebra.