Exact sequence in Hartshorne's proof of Clifford's theorem (Theorem IV.5.4) In Theorem 5.4 (Clifford) of chapter IV of Hartshorne we have an exact sequence:
$$0\rightarrow \mathscr{L}(D')\rightarrow \mathscr{L}(D)\oplus \mathscr{L}(E)\rightarrow \mathscr{L}(D+E-D')\rightarrow 0$$
where $D$ is a divisor $\neq 0,K$ (the canonical divisor) of degree $\geq 4$ and such that $\dim |D|=\frac{1}{2}\deg D$; $E\in |K-D|$ and $D'=D\cap E$ the largest divisor dominated by both $D$ and $E$.
Fix two points $P,Q\in X$ such that $P\in \operatorname{Supp} E$ and $Q\notin \operatorname{Supp} E$ and we choose $D$ such that $P,Q\in \operatorname{Supp} D$.
Why is it exact? It reads that the first map is addition, and the second is subtraction. What does it mean?
 A: Let's be really clear about the setup here. $\def\Supp{\operatorname{Supp}}\def\sL{\mathscr{L}}$

*

*$X$ is a curve

*$K$ is the canonical divisor

*$D$ is an effective special divisor, which is assumed to have $\deg D=2n$ and $\dim |D|=n$ for $n\geq 2$

*$E \in |K-D|$

*$P,Q\in X$ are two points such that $P\in\Supp E$ and $Q\notin\Supp E$

*$D'$ is the largest divisor dominated by $D$ and $E$. (That is, we can write $D$ as a sum of $D'$ and an effective divisor, and similarly with $E$.)

Since $\dim |D|\geq 2$ we can find an effective divisor in $|D|$ containing $P$ and $Q$, so we may assume $P,Q\in D$. Therefore $D'\neq D$, so $\deg D'<\deg D$, and since $P\in \Supp D'$, we have $0<\deg D'$.
Now we come to the bit you're having trouble with. Here's the direct quote from the text:

Next, by construction of $D'$, we have an exact sequence $$0\to \sL(D')\to\sL(D)\oplus\sL(E)\to \sL(D+E-D')\to 0$$ where we consider these as subspaces of the constant sheaf $\mathscr{K}$ on $X$, and the first map is addition, the second subtraction. (Think of $\sL(D)=\{f\in K(X)\mid (f)\geq -D\}$, cf. (II, 7.7).)

"Addition" isn't a great way to refer to the first map - what's going on is that we're sending an element $f\in\sL(D')$ to $(f,f)\in \sL(D)\oplus\sL(E)$, which works because if $(f)\geq -D'$, then as $D>D'$ and $E>D'$, $(f)\geq -D$ and $(f)\geq -E$.
The second map is actually subtraction - that is, given $(f,g)$ with $(f)\geq -D$ and $(g)\geq -E$, we look at the rational function $f-g$. It's not so hard to see that for a point $x\in X$, the order of the pole of $f-g$ at $x$ is at most the maximum of the order of $x$ in $D$ and the order of the $x$ in $E$. But this is the order of $x$ in $D+E-D'$: if $x$ has order $a$ in $D$ and order $b$ in $E$, then it has order $\min(a,b)$ in $D'$, and the result follows from noticing $\max(a,b)=a+b-\min(a,b)$.
Now let's check exactness. The first map is not so hard to check: $(f,f)=0$ iff $f=0$, so the first map is injective. It's image is clearly inside the kernel of the second map, and if $f\in\sL(D)$ is equal to $g\in\sL(E)$, then $(f)\geq -D$ and $(g)\geq -E$, so $(f)=(g)\geq -D'$ and so $(f,g)$ is in the image of the first map.
Exactness at the final term (i.e., surjectivity of subtraction) is actually unnecessary for the following argument (global sections is left-exact, so it sends an exact sequence of the form $0\to A\to B\to C$ to an exact sequence of the form $0\to F(A)\to F(B)\to F(C)$ which is enough to conclude the dimension of the middle term is at most the sum of the dimensions of the outer terms) and more difficult to prove, so we skip it.
