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.
Not sure if that helps you, but I remember it as follows:
Riemann-Roch for a curve $C$ (or Riemann surface if you prefer) says, that given some line bundle $\mathscr{L} \in \text{Pic}(C)$ (or some divisor), we have the equality
$$\chi(\mathscr{L}) = \text{deg}(\mathscr{L}) + \chi(\mathscr{O}_C).$$ Actually I prefer to state it as $$\text{deg}(\mathscr{L}) = \chi(\mathscr{L}) - \chi(\mathscr{O}_C)$$ since I can more easily remember the sign and to sort of use it as the definition of the degree as well (it somehow also seems more aesthetic to have all the Euler characteristics on one side).
Now under the usual mild assumptions, we have that $h^0(\mathscr{O}_C) = 1$ and we have that $h^1(\mathscr{O}_C) = g$ is the genus of the curve. Plugging that in the first equation yields the Riemann-Roch that you stated. So I guess what I am trying to say is that rephrasing both the left hand side and the $1 - g$ by some Euler characteristic makes it quite easy to remember.
For your other concerns, I would take a look at the comment of KReiser. I think the classical proof is what at least gives me the intuition.
Best Answer
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 $l(P)=1$.