Let $X$ be an elliptic curve over $k$, and $P \in X$. We have to prove that for $D=nP$ we have
$$
l(D)=\deg(D)=n
$$
If $D$ satisfies the above equation, and $E$ is a divisor such that $E \geq D$, then we have $l(E)=\deg(E)$ (cf. Fulton, Corollary 8.3.1). Hence it is enough to show that $l(P)=1$.
Clearly $l(P) > 0$, since $k \subset L(P)$. On the other hand, $l(P) > 1$ would imply that there exists $x \in k(X)$ such that $x$ has a simple pole at $P$, and no other poles. But this would imply that the map
$$
x: X \rightarrow \Bbb{P}^1
$$
is an isomorphism, which is a contradiction. Thus $l(P)=1$.
As has already been pointed out, it might be worthwhile learning a bit of algebraic geometry first but I'll try to answer this without going too deep. For most of this I will follow Silverman, as I think elliptic curves make divisors a lot more approachable than abitrary varieties or schemes.
Recall some basics of topology, every compact orientable 2-manifold looks like (is homeomorphic to) an $n$-torus, i.e. either a sphere or a doughnut shape with $n$ holes. This $n$ happens to be equal to the genus of this surface. How do we connect this to elliptic curves though?
Over $\mathbb{C}$, the solution set of an elliptic curve actually looks like a torus topologically (this is called the uniformisation theorem), as we quotient $\mathbb{C}$ by some lattice. Since this is simply a torus with 1 hole, the genus of an elliptic curve is $1$. We can also do this for hyperelliptic curves (although the genus will be different).
On to divisors:
Every place of the funtion field of an elliptic curve corresponds to a point on this curve, so we can just think of a divisor as a formal sum of points. The tiny catch is that this sum needs to be finite but that is perfectly fine.
For example, if $E: y^2=x^3+1$ over $\mathbb{C}$, then $Q=(2,3)$ and $R=(0,1)$ are points on $E$. We then have divisors
$D_1 = 2[R], D_2=-2[R]+[Q], D_3=[R]+[Q]$, and these form an abelian group under the obvious addition so $D_1+D_2=D_3$. (Forget about the group law on elliptic curves if you know about it). We also have the concept of the degree of a divisor, which is just the sum of the $n_i$. In the above the degrees of $D_1,D_2$ and $D_3$ are $2,-1$ and $1$ respectively.
Now consider a nonzero function $f$ in the function field of the curve. We say $v_P(f)=n$ if $f$ has a zero of order $n$ at $P$ (note $n$ can be negative if it has a pole instead. Checking for all poles and zeroes of this function, we can construct a divisor out of $f$, which is called $div(f)$.
For example, let $f=x-2$. Then $f=0$ if $x=2$ which occurs for the points $Q=(2,3)$ and $Q'=(2,-3)$ and we can easily see that these are simple zeroes. This means $v_Q(f)=v_{Q'}(f)=1$.
We can also see that $f$ does not have a pole at any point in the affine plane, however, elliptic curves really live in projective space and it turns out that $f$ has a pole of order $2$ at $\infty$. So $v_{\infty}(f)=-2$.
Having found all the zeroes and poles, we now see that the divisor of $f$ is $div(f)=[Q]+[Q']-2[\infty]$. Note that the degree of this is zero and this is always the case for divisors of functions.
For the Riemann Roch space $\mathcal{L}(D)$, we want to find all functions whose divisor is at least the negative of another divisor. For example, if we take $D=[Q]+[Q']$, then we are saying that we are only allowing functions which have at most simple poles at $Q$ and $Q'$ and nowhere else. It is an easy check that this forms a complex vector space.
Adjusting our idea above, $\dfrac{1}{x-2}$ lives in this space (as well as any scale multiple by the vector space structure), but no others do. Hence, we have $1$ basis element so $l(D) \geq 1$. We can also check that $\dfrac{x}{x-2}$ lives in here and that is it so $l(D)=2$. It is useful to note that since all nonconstant functions have poles, then if the degree of $D$ is negative then $l(D)=0$.
Now the Riemann Roch theorem can either be viewed as the definition for the genus, or if you already know this, it can be used to find $l(D)$, up to some correction term.
Now $\mathcal{K}$ is what is known as the canonical divisor and comes from a differential form which I won't attempt to explain here, but has degree $2g-2$.
So in the above, $\mathcal{K}-D$ has degree $-2$ which is negative hence $l(\mathcal{K}-D)=0$ and we can see that the Riemann Roch formula holds.
Best Answer
The idea is to prove that the map that you defined above (called the Abel-Jacobi map) $$\begin{align*} J \colon E(k) &\to \mathrm{Pic}^0(E)\\ P &\mapsto (P)-(O) \end{align*}$$
is a bijection, where $\mathrm{Pic}^0(E)$ is the subgroup of $\mathrm{Pic}(E)$ of elements of degree $0$. Then the group law of $E(k)$ will be the one that makes the map above an isomorphism of groups.
Remarks: In this context $\mathrm{Pic}(E)$ is just another notation for $\mathrm{Cl}(E)$. As $k$ may not be algebraically closed we have to recall that the degree of $D=\sum_{p\in E}n_p(P)$ is given by $\deg(D)=\sum_{p\in E}n_p[k(p):k]$ so we have $P\in E(k)$ if and only if $deg(P)=1$.
Injectivity of the map came from the fact that if $(P)-(Q)=\text{div}(f)$ then by replacing $D=(Q)$ on Riemann-Roch we get $h^0(Q)=2-g=1$ and hence $f\in H^0(E,Q)$ must be constant.
Over an algebraically closed field you can also proceed without R-R as here.
Surjectivity came from the fact that if $D\in \mathrm{Div}^0(E)$ then by R-R we have $h^0(D+(O))=1$ so if $f\in H^0(E,D+(O))$ is not constant we have $\text{div}(f)=-D-(O)+(P)$ for some $P$. As $\deg(D)=\deg(\mathrm{div}(f))=0$ we get $\deg(P)=\deg(O)=1$ hence $P\in E(k)$ and then $$P\mapsto (P)-(O)\sim D$$
Now to prove that this group law $\oplus$ coincides with the geometric group law (the one defined with lines) when $E$ is given by a Weierstrass equation $$E:Y^2Z=4X^3-aX^2-bZ^3$$ it's enough to notice that $P\oplus Q\oplus R = O$ $\iff$ $P,Q,R$ are coolinear $\iff$ there is a degree one homogeneous polynomial $F(X,Y,Z)$ with $V(F)=\{P,Q,R\}$ $\iff$ $\mathrm{div}(\frac{F}{Z})=(P)+(Q)+(R)-3(O)$ (notice that the intersection multiplicity between $V(Z)$ and $E$ is $3$, hence the $3(O)$ term) $\iff$ $P+Q+R=O$ with the addition descrived above. Notice that all the above its true when the set $\{P,Q,R\}$ degenerates with tangencies.