Xjyuk

Suppose that $C_1, C_2$ are two curves of genus $g geq 2$ defined over a number field $K$. Let $J_1, J_2$ respectively be their Jacobians. Suppose that $J_1, J_2$ are isogenous over $K$ and $C_1(K), C_2(K)$ are both non-empty, can $C_1(K), C_2(K)$ have different cardinalities?

For $g = 1$ and without the assumption that $C_i(K) ne emptyset$, the conclusion is obviously false. Take any elliptic curve $E$ over $mathbbQ$ of positive rank such that the $2$-Selmer group of $E$ is non-trivial, so that there is a genus one curve $C$ which represents a non-trivial $2$-Selmer element of $E$. Then $E$ is isomorphic to the Jacobian of $C$, and the Jacobian of $E$ is itself, so that $C,E$ have isogenous Jacobians but $E(mathbbQ)$ is by assumption infinite but $C(mathbbQ) = emptyset$.

Yes it is possible, already for $(K,g) = (bf Q,2)$, and
already with the first example of isogenous $C_1,C_2$ listed in the LMFDB:
curve
249.a.249.1,
$y^2 + (x^3+1) y = x^2 + x$, has one rational Weierstrass point, while curve
249.a.6723.1,
$y^2 + (x^3+1) y = -x^5 + x^3 + x^2 + 3x + 2$, has two, so the rational-point
counts don't even have the same parity. (The counts are known to be
$5$ and $6$ respectively.) Further examples are isogeny class
277.a
(two curves, each with a unique rational Weierstrass point,
but the total counts are $1$ and $5$), and isogeny class
644.a
(two curves, one with two rational points, the other not solvable over $bf R$
--- so this violates your assumption that each $C_i(K)$ is nonempty).
The first example where the counts do agree is the two curves in
isogeny class 294.a,
each with $4$ rational points.

By the way, for $g=1$ there are examples even with each $C_i(K)$ nonempty;
for example, take $K=bf Q$, let $C_1$ be either $X_0(11)$ or $X_1(11)$,
and let $C_2$ be the third elliptic curve of conductor $11$.
Then $C_1$ has five rational points and $C_1$ has only one.