The quotient by a maximal line subbundle is locally free

The twisted extension. By [F1] the morphism $\phi$ is injective, so $0\to {\mathcal{O}}_X(b)\to E\to F\to 0$ is exact with $F=E\mathrm{/}{\mathcal{O}}_X(b)$. Twisting by ${\mathcal{O}}_X(-b)$ and using ${\mathcal{O}}_X(b)\otimes {\mathcal{O}}_X(-b)\cong {\mathcal{O}}_X$ gives the exact sequence $0\to {\mathcal{O}}_X\to M\to F(-b)\to 0$ with $M:=E(-b)$; by [F1] $H^0(X,M)\mathrm{\ne }0$, while $H^0(X,M(-1))=H^0(X,E(-b-1))=0$ because $M(-1)\cong E(-b-1)$.

The quotient has no sections in the next twist. Twisting the sequence of step 1.1 by ${\mathcal{O}}_X(-1)$ gives the exact sequence $0\to {\mathcal{O}}_X(-1)\to M(-1)\to F(-b-1)\to 0$. Its long exact sequence [F3] reads $0=H^0(X,M(-1))\to H^0(X,F(-b-1))\to H^1(X,{\mathcal{O}}_X(-1))=0,$ because $H^0(X,{\mathcal{O}}_X(-1))=0$ as well; hence $H^0(X,F(-b-1))=0$.

$F(-b)$ is torsion-free. Suppose there are an open $U_0\subseteq X$, nonzero $m\in F(-b)(U_0)$ and nonzero $a\in {\mathcal{O}}_X(U_0)$ with $am=0$. Choose a point $x\in U_0$ with $m_x\mathrm{\ne }0$; this only uses that $m$ is a nonzero section, and the nonzero-germ locus is not asserted to be open. Choose one of the two standard charts $T$ containing $x$, then a principal affine open $U\subseteq U_0\cap T$ containing $x$. The restricted sections remain nonzero, and ${\mathcal{O}}_X(-1){\mathrm{\mid }}_U$ has a nowhere-vanishing frame $g$ by [F5, F11]. The assignment $s\mapsto s\otimes g$ is an isomorphism $F(-b){\mathrm{\mid }}_U\to F(-b-1){\mathrm{\mid }}_U$, so $m^{\mathrm{\prime }}:=m{\mathrm{\mid }}_U\otimes g\in F(-b-1)(U)$ is nonzero and $a{\mathrm{\mid }}_U{m}^{\mathrm{\prime }}=0$. Let $Z=\{y\in U:a_y\in {\mathfrak{m}}_y\}=V(a{\mathrm{\mid }}_U)$, the residue-zero locus. It is a proper closed subset of the integral curve $U$ because $a{\mathrm{\mid }}_U$ is nonzero [F11]; by [F11] it is finite and each point of $Z$ is closed in $X$. Thus $W:=X\setminus Z$ is open and $U\cup W=X$. On $U\cap W=U\setminus Z$, the residue of $a$ is nonzero at each point, so $a$ is a unit in every stalk and $m^{\mathrm{\prime }}{\mathrm{\mid }}_{U\cap W}=0$ by [F9]. The sections $m^{\mathrm{\prime }}$ over $U$ and $0$ over $W$ agree on the overlap and glue to a nonzero global section of $F(-b-1)$, contradicting step 2.1. Hence $F(-b)$ is torsion-free.

Local freeness. Fix a standard chart $U=\mathrm{Spec}A$, $A=k[t]$ or $k[u]$, and put $M:=F(-b)(U)$. By [F4], $F(-b)$ is coherent, hence quasi-coherent, so [F7] applies and $F(-b){\mathrm{\mid }}_U\cong \tilde{M}$, so $F(-b)(D(f))\cong M_f$ for every $f\in A$. By [F4] $F(-b)$ is coherent, hence of finite type, so every point $x\in U$ has an affine open $V_x\subseteq X$ with $F(-b){\mathrm{\mid }}_{V_x}\cong \widetilde{N_x}$ for a finitely generated module $N_x$; choosing $f_x\in A$ with $x\in D(f_x)\subseteq V_x\cap U$ [F7], putting $B_x:=\mathrm{\Gamma }(V_x,{\mathcal{O}}_X)$, the affine restriction isomorphism in [F7] identifies $M_{f_x}$ with $A_{f_x}{\otimes }_{B_x}{N}_x$, a finitely generated $A_{f_x}$-module whose generators are the images of any finite generating set of $N_x$. Since $U$ is affine, hence quasi-compact [F8], finitely many such $D(f_1),\dots ,D(f_k)$ cover $U$, so $f_1,\dots ,f_k$ generate the unit ideal of $A$ and each $M_{f_i}$ is finitely generated. Choose finitely many elements of $M$ whose images generate each $M_{f_i}$ and let $N\subseteq M$ be the submodule they generate. For every $z\in M\mathrm{/}N$, the vanishing $(M\mathrm{/}N{)}_{f_i}=0$ and [F7] give, for each $i$, a power $f_i^{n_i(z)}$ that annihilates this element $z$; the powers may depend on $z$. Since $(f_1,\dots ,f_k)=A$, also $(f_1^{n_1(z)},\dots ,f_k^{n_k(z)})=A$ (their generated ideal has the same radical as the unit ideal). Thus $z=0$, and $M\mathrm{/}N=0$: $M$ is a finitely generated $A$-module. By step 3.1 the module $M$ is torsion-free, and $A=k[t]$ (respectively $k[u]$) is a principal ideal domain [F5], so $M$ is free [F5]. As the two charts cover $X$, $F(-b)$ is finite locally free; twisting back by ${\mathcal{O}}_X(b)$, an equivalence by [F2], $F$ is finite locally free as well.

The rank. At each point $x\in X$, the stalk sequence from step 1.1 is exact and all three stalks are free after step 4.1. The surjection $E_x\to F_x$ splits because $F_x$ is free, so the ranks add: $r=1+\mathrm{rk}{F}_x$. Hence $F$ is locally free of rank $r-1$ everywhere.

Conclusion. The module $F$ is finite locally free of rank $r-1$ by steps 4.1 and 5.1. The Axiom of Choice enters only through the suppliers recorded in [F10].