The group law on elliptic curves comes from looking at the functions of geometric surfaces, and trying to understand the behaviour of intersections.
When we look at the coordinate ring of an elliptic curve, we have:
$$C = y^2 - x^3 - Ax - B$$
$$I = \langle C \rangle$$
$$K[V] = K[x, y] / I$$
As you can see, all polynomials that are multiple of the curve equation will also be equivalent to the $\bar{0}$ class in this ring.
We can also look at the points which disappear in this ring by taking the variety denoted $V(I) = \{ (a, b) \in K[x, y]: C(a, b) = 0 \}$. We also can look at the function field $K(V) = \{ f / g: f, g \in K[V] \}$, and begin to ask questions about where functions in this field disappear. That is for what points do they either become zero when $f(P) = 0$ or infinity when $g(P) = 0$.
By bezout's theorem, the sum of all these zeros (you can think of this as an intersection multiplicity) and infinities is the same as the curves degree, which in our case $\textrm{deg}(C) = 3$. For example given $P = (a, b)$, then the line $f = x - a$, will intersect $C$ at $P$ with a valuation of 1. Whereas when $f$ is a line tangent to the curve at $P$, then it can have a valuation of either 2 or even 3.
I won't go more into discrete valuation rings, but if you're interested to understand more then consult this sage file.
We can represent information about these curves using a divisor.
$$\textrm{div}(x - a) = [(x - a, b)] + [(x - a, -b)] + [\infty] - 3[\infty]$$
Let $C = y^2 - x^3 + 4x$ on $\mathbb{F}_{11}$. Then
$$\textrm{div}(y - 2x) = [(0, 0)] + 2[(2, 4)] - 3[\infty]$$
Which means $y - 2x$ intersects $C$ at the 3 points of $(0, 0), (2, 4), (2, 4)$. Since it is tangent, we count $(2, 4)$ twice. You could imagine slightly moving the line (or modify the curve roots slightly what we call perturbing), and you'd see that the single intersection will turn into 2 intersections.
As you can see when we take the zeros and infinities, the total sum of coefficients (what is called the divisor degree) becomes 0. Likewise the divisor of a function has a sum $\infty$.
$$\textrm{deg}(y - 2x) = 1 + 2 - 3 = 0$$
$$\textrm{sum}(y - 2x) = 1\cdot(0, 0) + 2\cdot(2, 4) - 3\cdot \infty = \infty$$
It is an interesting property of divisors of functions that their degree is always 0, and sum is always $\infty$.
Divisors of functions (called principal divisors) belong to a group called $\textrm{Prin}(E)$.
Zero degree divisors (which includes principal divisors) make up a group denoted by $\textrm{Div}^0(E)$.
The quotient of these is a group called the Picard group $\textrm{Pic}^0(E)$, and there is an isomorphic map
$$J : E(K) \rightarrow \textrm{Pic}^0(E)$$
$$J(P) = [P] - [\infty]$$
This is for elliptic curves, and it can be shown that the operation on the right inside the Picard group corresponds to a geometric interpretation (what you know as the elliptic curve group law).
Given ideal classes $\bar{D_1}, \bar{D_2} \in \textrm{Pic}^0(E)$, then we define equivalence $\bar{D_1} \sim \bar{D_2} \iff D_1 - D_2 \in \textrm{Prin}^0(E)$.
But actually with higher degree surfaces, this group law operation involving lines no longer works because any curve will intersect at n places on the curve.
It can be shown that every curve has a reduced divisor representation which corresponds to
$$[P_1] + \cdots + [P_g] - g[\infty]$$
It just so happens that elliptic curves have $g = 1$, and so we get the simple isomorphic map between the Picard group using the mapping $J$ above. In fact all genus 1 curves are isogenous to elliptic curves (hence their usefulness).
My aim with this isn't a rigorous treatment of the topic, but an intuitive introduction and some leads where to look. Lmk if that helps.
