One can take
$$
\begin{array}{llllll}
y^2 + (a_g x^g+...+a_0) y &=& x^{2g+1} &&& \text{ ($n=2g+1$ odd)} \cr
y^2 + (2cx_{g+1} +a_g x^g+...+a_0) y &=& -c^2x^{2g+2} &&& \text{ ($n=2g+2$ even).} \cr
\end{array}
$$
The divisor of the $y$ function is
$$
\text{div}(y) = n(0,0)-n(\infty),
$$
and so $D=(0,0)-(\infty)$ is an $n$-torsion point on the Jacobian.

Note that in both cases, completing the square on the left gives an equation $y^2=$ polynomial of degree $2g+1$, so the curve has a unique point at infinity. Also, the order of $D$ is exactly $n$, for otherwise $y$ would be (up to a constant) a power of a rational function which has a point of order $<n$ at $\infty$ and no other poles, and all such functions are polynomials in $x$.

P.S. We needed such `interesting' torsion points to construct elements of $K_2$ on hyperelliptic curves