In elliptic curve cryptography, how is "A dot A" computed?
You are correct that a line through a single point can intersect the curve in many other points. But you don't choose just any line. You choose the tangent through $A$. And that line will intersect the curve in exactly one point (which usually isn't, but may actually be $A$).

It is an interesting and useful exercise to actually do these computations with the formulae. Choose an elliptic curve, choose a point (not with $y$-coordinate $0$), find the curve's tangent line, then compute the remaining intersection point. (This is easy. You have two equations $Y^2 = X^3 + aX + B$ and $Y=\alpha X + \beta$. Eliminating $Y$, you get a cubic equation in $X$ and you know a double zero, namely the $x$-coordinate of the point you are trying to double.)