Kernels of surjections of vector bundles

Fix $x\in X$ and trivialize $E$ and $F$ near $x$, of ranks $n$ and $m$. Surjectivity on the stalk at $x$ implies surjectivity of the induced map of vector spaces over the residue field $\kappa (x)$. Its matrix therefore has an $m\times m$ minor nonzero at $x$; for $m=0$ use the empty minor, equal to $1$. After shrinking around $x$, the corresponding determinant is a unit. Permuting the basis of $E$ and applying the inverse of this minor to the chosen $m$ columns gives a right inverse $t:F\to E$ on this neighbourhood. In particular $m\le n$.

On that neighbourhood, $E=\ker f\oplus t(F)$: every $e$ equals $(e-tf(e))+tf(e)$, and the intersection is zero. More explicitly, the remaining $n-m$ basis vectors, corrected by subtracting their images under $tf$, form a basis of $\ker f$; their independent coordinates in the unselected positions prove independence and the displayed decomposition proves spanning. Thus the kernel is finite locally free of rank $n-m$, and $t$ is the required local splitting. Since $x$ was arbitrary, the conclusion holds on $X$.

For $s:E\to {\mathcal{O}}_X$, on a trivializing neighbourhood its image ideal is generated by the finitely many coefficients of $s$, so these local ideals define $V(I)$. At each point of its complement at least one coefficient is a unit after shrinking; hence $s{\mathrm{\mid }}_U$ is surjective. Restriction to an open set preserves kernels, because kernels are computed sectionwise. Applying the preceding construction with target rank $1$ proves the stated rank and local freeness. If $r=0$, $I=0$ and $U$ is empty.