The equation for $\# J_C(\mathbb F_p)$ that you quote contains a typo: they must have meant that $\# J_C(\mathbb F_p) = \frac 1 2 \# C(\mathbb F_{p^2}) + \frac 1 2 \# C(\mathbb F_p)^2 - p$, which is consistent with your example. That there is an equation like this is a direct consequence of the Grothendieck-Lefschetz trace formula: use that $H^i$ of an abelian variety is $\wedge^i$ of $H^1$, and that $H^1$ of a jacobian is isomorphic to $H^1$ of the curve.

The equation for $\# J_C(\mathbb F_p)$ that you quote contains a typo: they must have meant that $\# J_C(\mathbb F_p) = \frac 1 2 \# C(\mathbb F_{p^2}) + \frac 1 2 \# C(\mathbb F_p)^2 - p$, which is consistent with your example. That there is an equation like this is a direct consequence of the Grothendieck-Lefschetz trace formula: use that $H^i$ of an abelian variety is $\wedge^i$ of $H^1$, and that $H^1$ of a jacobian is isomorphic to $H^1$ of the curve.

If you think about it in terms of symmetric functions it's not hard to see that $\chi(n)$ is always going to be some universal expression in $\#C(\mathbb F_p)$ and $\# C(\mathbb F_{p^2})$. The coefficients in the characteristic polynomials are symmetric functions of the Frobenius eigenvalues, and are therefore expressible in terms of the power sums of the Frobenius eigenvalues. But knowing the power sums of the Frobenius eigenvalues is exactly the same as knowing $\#C(\mathbb F_{p^k})$ for all $k$. Since we are in genus two we only need $k=1,2$ to determine all Frobenius eigenvalues (using Poincaré duality/the functional equation).