Let $X$ be a projective variety over $\mathbb{C}$. If $F\subset E$ is a subbundle, then $F(-D)\subset E$ is a subsheaf, where $D$ is say an effective divisor on $X$ and $F(-D):=F\otimes \mathcal{O}_X(-D)$. Even though it's a subsheaf, why is $F(-D)$ not a subbundle of $E$ when $D\neq 0$? In the case $\mathcal{O}_X(-D)\subset \mathcal{O}_X$, it's not a subbundle because a line bundle does not have nonzero subbundles.
I know that generally, given locally free sheaves of $\mathcal{O}_X$-modules $\mathscr{G}$ and $\mathscr{H}$ such that $\mathscr{G}\subset\mathscr{H}$, though we have $0\rightarrow \mathscr{G}_x\rightarrow \mathscr H_x$ on the level of sheaves, tensoring with the residue field is not exact, and so we only have $$\cdots\rightarrow \operatorname{Tor}^1(k(x),\mathscr H_x/\mathscr G_x)\rightarrow G_x\rightarrow H_x\rightarrow\cdots$$ where $G_x=\mathscr{G}_x\otimes k(x)$ and $H_x=\mathscr{H}_x\otimes k(x)$. So if $\mathscr H_x/\mathscr G_x$ is not flat, we don't have that $G$ is a subbundle of $H$. I would also like to see how this fails explicitly in the case $F(-D)\subset E$ and $D\neq 0$. And how can I show that a line bundle doesn't have nontrivial subbundles using this setup?