Simple exact sequences of vector bundles I've come across some simple exact sequences of vector bundles that make manifest some basic confusions I have. These questions may be quite intertwined, in ways that my limited understanding obscures, so I thought I would ask them together to form a slightly broader `simple vector bundle exact sequences' question. Below are three questions.
1) Given a divisor $D$ of a space $X$, I have seen the following short exact sequence,
$$ 0 \to \mathcal{O}_X(-D) \to \mathcal{O}_X \to \mathcal{O}_X|_D \to 0 \,. $$
I would have thought that an exact sequence between vector bundles would imply exact sequences between the fibres (which are vector spaces), so that the alternating sum of the dimensions of these fibres would be zero. Hence I'm confused as to how we can have three line bundles in an exact sequence. I would also like to understand what are the morphisms involved (and how to think of the restriction map as a morphism).
2) Given two divisors $D$ and $F$ (perhaps with some extra assumptions), I have seen the following short exact sequence,
$$ 0 \to \mathcal{O}_{D+F} \to \mathcal{O}_D\oplus\mathcal{O}_F \to \mathcal{O}_{D\cap F} \to 0 \,. $$
In this case, I would just like to understand what are the morphisms used here and how best to think about them.
3) My third question concerns the appearance of `trivial' bundles in a general exact sequence. Let $V$ and $V'$ be vector bundles over $X$, and let $S\subset X$ be a subspace of $X$. Suppose we have an exact sequence of vector bundles,
$$ 0 \to V \to V' \to \mathcal{O}_X|_S \to 0 \,.$$
The final vector bundle is in some sense trivial. Does this have special implications for the relationship between $V$ and $V'$, i.e. can we claim more than $\mathcal{O}_X|_S = V'/V$?
 A: For (1), note that $\mathcal{O}_D$ is in fact shorthand for $i_\ast \mathcal{O}_D$, the pushforward of the structure sheaf $\mathcal{O}_D$ by the inclusion $D \hookrightarrow X$.  In particular, while $\mathcal{O}_D$ is a (trivial) line bundle on $D$, it's not the case that $i_\ast \mathcal{O}_D$ is a line bundle on $X$.  For instance, if $x \in X$ is any point with $x \not \in D$ then we have $\mathcal{O}_{D, x} = 0$.
For (2), note that if we have a function on the space $D \cup F$, we restrict to $D$, giving us an element of $\mathcal{O}_D$, and similarly we can restrict to $F$, giving an element of $\mathcal{O}_F$.  That's what the first map $\mathcal{O}_{D+F} \to \mathcal{O}_D \oplus \mathcal{O}_F$ is.  (By the way, neither of these is a line bundle on $X$, and $\mathcal{O}_D \oplus \mathcal{O}_F$ -- which is again shorthand, namely for $(i_1)_\ast \mathcal{O}_D \oplus (i_2)_\ast \mathcal{O}_F$ -- isn't a line bundle on anything.)
The map $\mathcal{O}_D \oplus \mathcal{O}_F$ takes a pair $(f,g)$, where $f$ is a function on $D$ and $g$ is a function on $F$, restricts both of them to $D\cap F$, and subtracts them.  Obviously if you got $f$ and $g$ by restricting some original function then this will vanish, since both restrictions will be the same.
For (3), $\mathcal{O}_X|_S$ is the same thing as $\mathcal{O}_S$, and the restrictions of the other things are vector bundles too, so the presence of $X$ is totally superfluous here (unless you know something more about $X$ that rules out a lot of the bundles on $S$ from extending to $X$).  
