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Basically, the problem is that there are really five intersection points with your line (two of them are complex non-real). If you could make a group in the same way as for elliptic curves, then this …

You get an open subset of the Jacobian by looking at points "in general position", i.e., points represented by divisors of the form $(P)+(P')-2(\infty)$, where $\infty$ denotes the point at infinity, …

By now there is a fairly rich literature on computing the set of rational points on curves of higher genus, see for example my survey paper on "Rational points on curves". What one can do for your co …

It turns out that $C(K) = C(\mathbb Q) = \{\infty_+, \infty_-, (0,1), (0,-1), (1,1), (1,-1)\}$. To see this, consider a point $P \in C(K)$ and write $\bar{P}$ for its image under the nontrivial automo …

The relevant information can be obtained from a regular model of the curve over ${\mathbb Z}_p$. Such a model can be computed by repeatedly blowing up points or components of the special fiber that ar …

I would suggest that you work with the Kummer surface $K$ of $J$ instead of using Mumford coordinates. The advantage is that $K$ is a quartic surface in $\mathbb P^3$; in the case you are considering …