Riemann-Roch analysis of point divisor ring on smooth genus 3 Riemann surface Let $C=C_4\subset\mathbb{P}^2$ be the smooth genus 3 Riemann surface given by a quartic curve.
Let $P\in C$ be a point, and $D=P$ the divisor given by the point $P$.
Let $R(D)=\bigoplus_{n\geqslant0}\mathcal{L}(nD)$ be the graded ring associated with the divisor $D$.
We know that a canonical divisor $k$ on $C$ has degree $2g-2=4$ and that it is thus linearly equivalent to a hyperplane divisor $H=H_L$ for any line $L\subset\mathbb{P}^2$.
So by Riemann-Roch, we know that $\ell(nD)=n-2$ for $n\geqslant5$, since then $\deg(k-nD)<0$ thus $\ell(k-nD)=0$.
My question then is how do we calculate $\ell(nD)$ for $n=2,3,4$?
The cases $n=0,1$ give us $\ell(nD)=1$, and we know that $\ell(nD)$ is non-decreasing.
If we can further show that $\ell(nD),\ell(k-nD)>0$ then we can use Clifford's theorem to obtain some bounds, but this still gives us a few possible options.
 A: There is no uniform answer to your question: the  numbers $l(3P)$ and $l(4P)$   you are asking about  depend on the curve $C$ and on the point $P$.  
a) We always have  $l(2P)=1$
Indeed, $1\leq l(2P)\leq 2$ for a curve of positive genus while for genus $g\geq 2$ the existence of a point $P$  with  $l(2P)=2$ characterizes hyperelliptic curves.
However a smooth plane  curve  is never hyperelliptic and in our case we thus have $l(2P)=1$.
b) We  have  $l(4P)=2$ or $3$
Riemann-Roch  and Serre duality  yield $l(4P)=2+l(K-4P)$.
Since, as you mentioned, the canonical divisors are exactly the hyperplane sections we have   $l(K-4P)=1$ or $0$ according as the tangent to $C$ at $P$ cuts  $C$ in $4$ points or in less than $4$ points.
Of course generically this tangent cuts $C$ in two points , so that generically  $l(K-4P)=0$ and thus $l(4P)=2$.
However it can happen that the tangent at $P$ cuts $C$ in $4$ points: this is the case for $P=[1:e^{\frac {i\pi}{4}}:0]$ on the Fermat curve $x^4+y^4+z^4=0$ .
In such cases $l(4P)=3$.
c)    We  have  $l(3P)=1$ or $2$
 i) If $l(4P)=3$ we necessarily have $l(3P)=2$, since $l(4P)-1=2\leq l(3P)\leq l(2P)+1=2$.
ii) If $l(4P)=2$ we have $l(3P)=1$ or $2$: generically $l(3P)=1$ but $l(3P)=2$ iff $P$ is an inflexion point of $C$.
(Recall that a smooth quartic has between $1$ and $24$ inflexion points.
More precisely it has 24 inflexion points if we count them with a suitable multiplicity)
