Another use of embedding a curve into its Jacobian is to apply the Mordell-Weil Sieve'. Suppose $k = \mathbb Q$ for simplicity and that you don't know a rational point on $X$, but you know a rational divisor (class) $D$ of degree 1 on $X$. Then you can use $D$ to define an embedding $\iota$ of $X$ into its Jacobian $J$. Now assume in addition that $J({\mathbb Q})$ J(\mathbb Q)$is known explicitly. Then for every prime$p$(of good reduction, say), you can consider the images of$X({\mathbb F}_p)$X(\mathbb F_p)$ and of $J({\mathbb Q})$J(\mathbb Q)$in$J({\mathbb F}p)$J(\mathbb F_p)$ (the first under $\iota$, the second by reduction mod $p$). Clearly, $X({\mathbb Q})$X(\mathbb Q)$has to map into the intersection of these two. Now instead of considering one prime, we can consider all primes in a finite set$S$and look at the product of$\prod{\iota(X(\mathbb F_p))$over all$p \in S} \iota(X({\mathbb F}p}))$S$ and the image of $J({\mathbb Q})$ J(\mathbb Q)$in the product of$\prod{J(\mathbb F_p)$over all$p \in S} J({\mathbb F}_p)$S$.
Now for a suitable choice of $S$ it may be the case that the two sets are disjoint, which then proves that $X$ has no rational points. There are good reasons to believe that it is always possible to prove that $X({\mathbb Q})$ is empty in this way (if it is empty). See this paper.
Another use of embedding a curve into its Jacobian is to apply the Mordell-Weil Sieve'. Suppose $k = \mathbb Q$ for simplicity and that you don't know a rational point on $X$, but you know a rational divisor (class) $D$ of degree 1 on $X$. Then you can use $D$ to define an embedding $\iota$ of $X$ into its Jacobian $J$. Now assume in addition that $J({\mathbb Q})$ is known explicitly. Then for every prime $p$ (of good reduction, say), you can consider the images of$X({\mathbb F}_p)$and of$J({\mathbb Q})$in$J({\mathbb F}p)$ (the first under $\iota$, the second by reduction mod $p$). Clearly,$X({\mathbb Q})$has to map into the intersection of these two. Now instead of considering one prime, we can consider all primes in a finite set $S$ and look at$\prod{p \in S} \iota(X({\mathbb F}p}))$and the image of $J({\mathbb Q})$ in$\prod{p \in S} J({\mathbb F}_p)$.
Now for a suitable choice of $S$ it may be the case that the two sets are disjoint, which then proves that $X$ has no rational points. There are good reasons to believe that it is always possible to prove that $X({\mathbb Q})$` is empty in this way (if it is empty). See this paper.