Regularity gives generation, multiplication, and vanishing

Extend the field to $k(u)$ if necessary. By faithful flatness and [F1], both vanishing and surjectivity descend; generation descends by applying faithful flatness to the evaluation cokernel. We may therefore assume $k$ infinite. There is a hyperplane $H$ avoiding the finitely many associated points of $F$: hyperplanes containing a fixed associated point form a proper linear subset of the dual projective space, and a finite union of these subsets cannot exhaust its rational points over an infinite field. Multiplication by its equation gives $0\to F(-1)\to F\to F_H\to 0$.

Induct on $n$, the case $n=0$ being immediate. The long exact sequence at twist $m-i$ gives $H^i(F_H(m-i))=0$, since its adjacent groups $H^i(F(m-i))$ and $H^{i+1}(F(m-i-1))$ vanish. Thus $F_H$ is $m$-regular. For each $i>0$, start at $t=m-i$ and induct on $t$ using $H^i(F(t-1))\to H^i(F(t))\to H^i(F_H(t))$; the first term vanishes by the induction on $t$, and the last by induction on $n$. This proves all the stated vanishings.

For $t\ge m$, $H^1(F(t-1))=0$ makes $H^0(F(t))\to H^0(F_H(t))$ surjective. The restriction $H^0(\mathcal{O}(1))\to H^0({\mathcal{O}}_H(1))$ is also surjective. Induction on $n$ makes the multiplication on $H$ surjective. Consequently every section of $F(t+1)$ is the sum of a product of a linear form with a section of $F(t)$ and a section in the kernel of restriction to $H$. This kernel is multiplication by the equation of $H$ on $H^0(F(t))$, and is itself in the image of multiplication. The desired multiplication is surjective.

Iterating multiplication shows that $H^0(F(t))\otimes H^0(\mathcal{O}(a))\to H^0(F(t+a))$ is surjective for $a\ge 0$. For sufficiently large $a$, $F(t+a)$ is globally generated by [F1]. At a stalk, the products of sections all lie in the image of $H^0(F(t))\otimes \mathcal{O}(a)\to F(t+a)$; this image is therefore the whole stalk. Tensoring by the inverse invertible twist gives generation of $F(t)$. Field descent in step 1.1 completes the argument over arbitrary $k$.