Exact sequence in Beauville's "Complex Algebraic Surfaces" On page 3 of Beauville's book (Lemma I.5) he takes two curves $C$ and $C'$ in a surface $S$ an takes global sections $s\in H^0(S,\mathcal{O}_S(C))$ and $s'\in H^0(S,\mathcal{O}_S(C'))$. In a recent post, I was explained that you can take $s$ and $s'$ to be 1. Beauville then writes the sequence
$0\longrightarrow\mathcal{O}_S(-C-C') \xrightarrow{(s',-s)}\mathcal{O}_S(-C)\oplus \mathcal{O}_S(-C')\xrightarrow{(s,s')}\mathcal{O}_S\longrightarrow\mathcal{O}_{C\cap C'}\longrightarrow 0.$
I assume that if $\mathcal{O}_S(-C)$ (resp. $\mathcal{O}_S(-C')$)  is generated on an open set $U_\alpha$ by $f_\alpha$ (resp. $f_\alpha'$), then it makes sense that the first map in the sequence takes $f_\alpha f_\alpha'$ to $(f_\alpha,-f_\alpha')$, but I don't see how this is consistent with the notation $(s',-s)$, especially since $s$ and $s'$ can be taken to be 1! 
Where am I wrong here or not understanding something correctly?
 A: Let me just explain the morphism of sheaves $\mathcal{O}_S(-C-C') \xrightarrow{s'}\mathcal{O}_S(-C)$, the rest being similar.  
On  $U_\alpha $ the divisor $(-C-C')\mid U_\alpha$ is given by the holomorphic function $f_\alpha f'_\alpha$ and the section $s'\in \Gamma (U _\alpha ,\mathcal{O}_S(C'))$ is given by a meromorphic function function $\frac {S'_\alpha}{f'_\alpha }$ where $S'_\alpha $ is holomorphic on $U_\alpha$.
The morphism $\mathcal{O}_S(-C-C') \xrightarrow{s'}\mathcal{O}_S(-C)$ then sends the section $\frac {h_\alpha}{f_\alpha f'_\alpha}\in \Gamma (U_\alpha, \mathcal{O}_S(-C-C') )$ to the section $\frac {h_\alpha S'_\alpha}{f_\alpha }\in \Gamma (U_\alpha, \mathcal{O}_S(-C) )$.  
In order to reassure you, let us take for an example the canonical section  $s'_0=1=\frac {f'_\alpha}{ f'_\alpha}\in \Gamma(U_\alpha,\mathcal O_C')$ i.e. $S'_\alpha=f'_ \alpha$, about which you worry in your question.
The morphism $\mathcal{O}_S(-C-C') \xrightarrow{s'_0}\mathcal{O}_S(-C)$ is then given on $U_\alpha$ by $$\Gamma (U_\alpha, \mathcal{O}_S(-C-C') )\to \Gamma (U_\alpha, \mathcal{O}_S(-C) ):\frac {h_\alpha}{f_\alpha f'_\alpha}\mapsto \frac {h_\alpha f'_\alpha}{f_\alpha }$$
