The jump l(D+p) - l(D) ranges from zero to the residue degree

Set-up. By [F5] one has $L(D)\subseteq L(D+p)$ and $0\le l(D+p)-l(D)\le [\kappa (p):k]$ for every divisor $D$ and closed point $p$ on $C$; by [F4] a rational function $f\in k(C{)}^{\times }$ lies in $L(D)$ if and only if ${\mathrm{ord}}_x(f)+n_x\ge 0$ for every closed point $x$, where $n_x$ is the coefficient of $D$. On ${\mathbb{P}}_k^1$ every nonzero rational function is a quotient $A\mathrm{/}B$ of nonzero polynomials (Divisors on the projective line are classified by degree, For every field $F$, $F[x]$ is a unique factorisation domain), and by [F2] and [F3] its orders are ${\mathrm{ord}}_x(A\mathrm{/}B)={\mathrm{ord}}_x(A)-{\mathrm{ord}}_x(B)$ at every closed point, with ${\mathrm{ord}}_{\mathrm{\infty }}(A\mathrm{/}B)=\mathrm{deg}B-\mathrm{deg}A$ for $A\mathrm{/}B$ in lowest terms.

Case (1): $D=-2[q]$, $p=[a]$, $q=[0]$, $a\mathrm{\ne }0$. A rational function $f=A\mathrm{/}B$ in lowest terms lies in $L(D)$ exactly when ${\mathrm{ord}}_q(f)\ge 2$ and ${\mathrm{ord}}_x(f)\ge 0$ for every $x\mathrm{\ne }q$: the second condition says $f$ has no poles at all, so $B$ is constant and $f=A$ is a polynomial, while the first says the polynomial $A$ has a zero of order at least $2$ at $q$; the condition at infinity ${\mathrm{ord}}_{\mathrm{\infty }}(A)=-\mathrm{deg}A\ge 0$ forces $\mathrm{deg}A=0$, and then ${\mathrm{ord}}_q(A)=0<2$. Hence $L(D)=0$. The same argument with the single change ${\mathrm{ord}}_a(f)\ge -1$, applied to $L(D+p)=L(-2[q]+[a])$, says $f=A\mathrm{/}B$ is in $L(D+p)$ only if $f=A\mathrm{/}(t-a)$ with $\mathrm{deg}A\le 1$ and $A$ divisible by $t^2$ at $q$, which is impossible for a nonzero polynomial of degree at most one; hence $L(D+p)=0$ and the jump is $0$.

Case (2): $D=-[q]$, $p=[a]$, $a\mathrm{\ne }q$. Here $f\in L(D)$ requires ${\mathrm{ord}}_q(f)\ge 1$ and ${\mathrm{ord}}_x(f)\ge 0$ for $x\mathrm{\ne }q$: again $f=A$ is a polynomial with a zero at $q$, and ${\mathrm{ord}}_{\mathrm{\infty }}(A)=-\mathrm{deg}A\ge 0$ forces $\mathrm{deg}A=0$, contradicting the zero at $q$; so $L(D)=0$. For $D+p=[a]-[q]$ the conditions are ${\mathrm{ord}}_q(f)\ge 1$, ${\mathrm{ord}}_a(f)\ge -1$ and ${\mathrm{ord}}_x(f)\ge 0$ for all other $x$: writing $f=A\mathrm{/}(t-a)$ with $A\in k[t]$, the condition at infinity is ${\mathrm{ord}}_{\mathrm{\infty }}(f)=1-\mathrm{deg}A\ge 0$, so $\mathrm{deg}A\le 1$, and the condition at $q$ is $A(q)=0$; hence $A=c(t-q)$ and $L(D+p)$ is the one-dimensional span of $(t-q)\mathrm{/}(t-a)$, a nonzero function with ${\mathrm{ord}}_q=1$, ${\mathrm{ord}}_a=-1$, all other orders zero, so $\mathrm{div}((t-q)\mathrm{/}(t-a))=[q]-[a]=-(D+p)$. The jump is $1$, and for the rational point $p=[a]$ one has $[\kappa (p):k]=1$.

Case (3): $k=\mathbb{R}$, $D=0$, $p=V(t^2+1)$. By [F1] the polynomial $t^2+1$ is monic irreducible of degree two with closed point $p$ of residue degree $[\kappa (p):k]=2$ and $\mathrm{div}(t^2+1)=[p]-2[\mathrm{\infty }]$, so $[p]$ is linearly equivalent to $2[\mathrm{\infty }]$. The space $L(0)$ consists of the rational functions with ${\mathrm{ord}}_x(f)\ge 0$ at every closed point: these are the polynomials (since the denominator of a reduced fraction would give a pole) with ${\mathrm{ord}}_{\mathrm{\infty }}(A)=-\mathrm{deg}A\ge 0$, i.e. the constants, so $l(0)=1$. For $L(p)$: a reduced fraction $f=A\mathrm{/}B$ with all orders at least $0$ except possibly ${\mathrm{ord}}_p(f)\ge -1$ has $B=1$ or $B=t^2+1$, so $f=A\mathrm{/}(t^2+1)$ with $A\in \mathbb{R}[t]$, and the condition at infinity is ${\mathrm{ord}}_{\mathrm{\infty }}(f)=2-\mathrm{deg}A\ge 0$, i.e. $\mathrm{deg}A\le 2$. Hence $L(p)=\{A\mathrm{/}(t^2+1):\mathrm{deg}A\le 2\}$, spanned by $1\mathrm{/}(t^2+1)$, $t\mathrm{/}(t^2+1)$, $t^2\mathrm{/}(t^2+1)$: these three are linearly independent because clearing the denominator turns a relation into a polynomial identity of degree at most two. Therefore $l(p)=3$ and the jump is $2=[\kappa (p):k]$, realized over the non-algebraically-closed field $\mathbb{R}$.

Case (4): $k=\mathbb{R}$, $D=-2[\mathrm{\infty }]$, $p=V(t^2+1)$. Here $f\in L(D)$ requires ${\mathrm{ord}}_{\mathrm{\infty }}(f)\ge 2$ and ${\mathrm{ord}}_x(f)\ge 0$ for every $x\mathrm{\ne }\mathrm{\infty }$: as in step 1.2 this forces $f=A$ to be a polynomial with $-\mathrm{deg}A\ge 2$, impossible for a nonzero polynomial, so $L(D)=0$. In $L(D+p)$ the conditions are ${\mathrm{ord}}_{\mathrm{\infty }}(f)\ge 2$, ${\mathrm{ord}}_p(f)\ge -1$ and ${\mathrm{ord}}_x(f)\ge 0$ for all other $x$: a reduced fraction with no pole outside $p$ is of the form $f=A\mathrm{/}(t^2+1)$ with $A\in \mathbb{R}[t]$, and $2-\mathrm{deg}A={\mathrm{ord}}_{\mathrm{\infty }}(f)\ge 2$ forces $\mathrm{deg}A=0$; hence $L(D+p)$ is the one-dimensional span of $1\mathrm{/}(t^2+1)$, whose orders are $+2$ at $\mathrm{\infty }$ and $-1$ at $p$, so the jump is $1$ while $[\kappa (p):k]=2$. This is the intermediate value: the jump is neither $0$ nor the full residue degree.

Assembly and choice accounting. Steps 1.2-2.1 exhibit jumps $0$, $1$ and $2$ on ${\mathbb{P}}_k^1$, and in particular the extreme value $[\kappa (p):k]$ occurs for $[\kappa (p):k]=1$ in case (2) and for $[\kappa (p):k]=2$ in case (3), while case (4) gives the intermediate value $1$ in residue degree $2$; the general inequality $0\le l(D+p)-l(D)\le [\kappa (p):k]$ is [F5]. The sheaf restatement of case (3), $\mathcal{O}(p)\cong \mathcal{O}(2)$, follows from the current Cartier/Picard route [F4]. The Axiom of Choice is inherited only through the suppliers of [F1]-[F5], as recorded in [F6]; no infinite selection is made above, since the spanning functions are exhibited by formulas.