My question: how can I describe, fiberwise, this homomorphism $\mathcal{O}(n)\rightarrow \mathcal{O}(m)$ in terms of the degree-$(m-n)$ polynomial, when $n<0$?
Answer: let $C$ be the projective line. There is by Hartshorne, Ex II.5.1 isomorphisms
$$Hom_{\mathcal{O}_C}(\mathcal{O}(m), \mathcal{O}(n)) \cong
Hom_{\mathcal{O}_C}(\mathcal{O}_C, Hom(\mathcal{O}(m), \mathcal{O}(n))) \cong $$
$$Hom_{\mathcal{O}_C}(\mathcal{O}_C, \mathcal{O}(m)^*\otimes \mathcal{O}(n))\cong Hom_{\mathcal{O}_C}(\mathcal{O}_C, \mathcal{O}(n-m)) \cong H^0(C, \mathcal{O}(n-m)).$$
Here you should check that to give a map of $\mathcal{O}_C$-modules
$$s: \mathcal{O}_C \rightarrow \mathcal{O}(n-m)$$
is equivalent to give a global section of $\mathcal{O}(n-m)$ which is equivalent to giving a homogeneous polynomial of degree $n-m$ in $x_0,x_1$.
If $n-m < 0$ there are no such maps.
If $n-m \geq 0$ the vector space of such maps has dimension $n-m+1$, and given a map $s$ it corresponds 1-1 to a homogeneous polynomial
$$s(x_0,x_1) \in \mathbb{C}[x_0,x_1]^{n-m}.$$
Example: A map $s:\mathcal{O}(m) \rightarrow \mathcal{O}(n)$
corresponds 1-1 to a homogeneous polynomial $s(x_0,x_n)$ of degree $n-m$ from the following argument:
We get a map of graded modules
$$ \phi: \mathbb{C}[x_0,x_1](m)\rightarrow \mathbb{C}[x_0,x_1](n)$$
defined by
$$\phi(f(x_0,x_1)):=s(x_0,x_1)f(x_0,x_1).$$
Sheafifying this map we get an induced map
$$s:\mathcal{O}(m) \rightarrow \mathcal{O}(n)$$
of $\mathcal{O}_C$-modules. This construction gives all such maps by the above argument.
Example: For any $n,m$ with $n-m \geq 0$ you get a vector space of dimension $n-m+1$ of such maps.
The fiber: The induced map at the fiber at $x:=(t-\alpha) \in C(\mathbb{C})$: Let $t:=x_1/x_0$. We get the following map $\phi_x$ between fibers $\mathcal{O}(m)(x)\cong \mathbb{C}x_0^m:=\mathbb{C}[t]/(t-\alpha)x_0^m$ and $\mathcal{O}(n)(x)$:
$$\phi_x: \mathbb{C}x_0^m \rightarrow \mathbb{C}x_0^n$$
defined by
$$\phi(x)(ux_0^m):=g(\alpha)ux_0^n$$
where
$s(x_0,x_1)=g(t)x_0^{n-m}$.
Here the point $x:=(a_0:a_1)$ has $a_0 \neq 0$ and by definition $x=(1: a_1/a_0)=(1:\alpha)$. There is an isomorphism
$$\mathcal{O}(m)(D(x_0)) \cong \mathbb{C}[t]x_0^m$$
and the point $x$ corresponds 1-1 to the maximal ideal $(t-\alpha) \subseteq \mathbb{C}[t]$ with $t:=x_1/x_0$. The fiber at $x$ is by definition
$$\mathcal{O}(m)(x)\cong \mathbb{C}[t]/(t-\alpha) x_0^m \cong\mathbb{C}x_0^m.$$
A generalization: Similar results hold for any projective scheme $X \subseteq \mathbb{P}^n_{A}$:
There are isomorphisms
$$Hom_{\mathcal{O}_X}(\mathcal{O}(m), \mathcal{O}(n)) \cong Hom_{\mathcal{O}_X}(\mathcal{O}_X, \mathcal{O}(n-m))$$
$$ \cong H^0(X, \mathcal{O}(n-m)).$$
Hence for projective space $\mathbb{P}^n_k$ you get an equality
$$Hom_{\mathcal{O}}(\mathcal{O}(m), \mathcal{O}(n)) \cong k[x_0,..,x_n]^{n-m}.$$
Hence a map of sheaves
$$s: \mathcal{O}(m) \rightarrow \mathcal{O}(n)$$
is in 1-1 correspondence with a homogeneous polynomial
$$s(x_0,..,x_n) \in \mathbb{C}[x_0,..,x_n]^{n-m}$$
of degree $n-m$.
Note: Usually we write $k[x_i]_d$ instead of $k[x_i]^d$. The notation
$$k[x_i]^d :=\oplus_{i=1}^d k[x_i]e_i$$
usually means "the direct sum of $d$ copies of $k[x_i]$".
Note moreover: There is an exercise (Ex.II.5.9 in Hartshorne) classifying coherent sheaves on a projective scheme $X:=Proj(S) \subseteq \mathbb{P}^n_A$ with $A$ a finitely generated algebra over a field $k$. Assume $S$ is generated by $S_1$ as $A$-algebra and for a given graded $S$-module $M$, let
$$M_{\geq d}:=\oplus_{n \geq d} M_n.$$
Two graded $S$-modules $M,N$ are equivalent iff there is an integer $d\geq 0$ and an isomorphism $M_{\geq d} \cong N_{\geq d}$. A module $M$ is "quasi finitely generated" iff it is equivalent to a finitely generated module.
The functor $F(M):=\tilde{M}$ gives an equivalence between the category of coherent sheaves on $X$ and the category of quasi finitely generated $S$-modules modulo this equivalence.
Example: In particular if $A:=\mathbb{C}$ is the field of complex numbers and $E$ is a coherent analytic sheaf on $X$, there is a quasi finitely generated $S$-module $M$ and an "isomorphism" $E \cong \tilde{M}^s$ where $(-)^s$ is the analytification functor described in
Smooth algebraic varieties are complex manifolds
Hence if $L^s, E^s$ are coherent analytic sheaves on $X$ with $L^s$ a line bundle, there is an isomorphism
$$Hom_{\mathcal{O}_X}(L,E) \cong Hom_{\mathcal{O}_X}(\mathcal{O}_X, E\otimes L^*) \cong H^0(X, E\otimes L^*) \cong $$
$$Hom_{\mathcal{O}_X^s}(\mathcal{O}_X^s, E^s \otimes (L^*)^s) \cong H^0(X^s, E^s\otimes (L^*)^s).$$
Hence you can calculate the maps from $L^s$ to $E^s$ using $L$ and $E$. Hence if $L^s,E^s$ are coherent analytic sheaves on $X^s$ and if $L^s$ is invertible, the set of maps $L^s \rightarrow E^s$ is finite dimensional: Since $E^s\otimes (L^*)^s$ is coherent it follows from HH.II.5.19:
The vector space
$$\Gamma(X^s, E^s\otimes (L^*)^s):=H^0(X^s, E^s\otimes (L^*)^s)\cong H^0(X, E\otimes L^*)$$
is finite dimensional. This is Theorem AppB.2.1 in Hartshorne.
