The set $J(C)_{\Theta}[n]$ has the structure of a smooth irreducible algebraic curve, and the restriction of $J(C)\xrightarrow{\times n } J(C)$ to $C$ defines a morphism $J(C)_{\Theta}[n]\rightarrow C$ which is a finite etale cover with Galois group $J(C)[n]$. I don't think it is good to think of $J(C)_{\Theta}[n]$ as generalized torsion points. I think it's better to think of $J(C)_{\Theta}[n]$ as a curve which is an unramified cover of $C$. For example, while $J(C)[n]$ has $n^4$ points over $\bar{\mathbb{F}}_q$ (when $q$ is coprime to $n$), $J(C)_{\Theta}[n]$ has infinitely many $\bar{\mathbb{F}}_q$-points. Also by Riemann-Hurwitz the genus of $J(C)_{\Theta}[n]$ is $n^4+1$, so the Hasse-Weil bound tells you something about its $\mathbb{F}_q$-points.

For your second question about $J(C)_{\Theta}[m] \cap J(C)_{\Theta}[m]$, I will interpret this as a scheme theoretic intersection. (In other words, incorporating multiplicities.) We may calculate this using some facts about intersection theory on surfaces and line bundles on abelian varieties. As noted before, the curve $J(C)_{\Theta}[m]$ is the pullback of the divisor $C\subset J(C)$ (embedded using your implicitly chosen point $\mathcal{O}$). By the theorem of the square, we have the following linear equivalence of divisors on $J(C)$: $$J(C)_{\Theta}[n] \sim \frac{n^2+n}{2} C + \frac{n^2-n}{2} [-1]^*C.$$ Here $[-1]^*C$ is the image of $C$ under the $[-1]$ map. Since $[-1]^*C$ is algebraically equivalent to $C$ (even linearly equivalent if you choose $\mathcal{O}$ to be a Weierstrass point), we conclude that $$ (J_{\Theta}(C)[m],J_{\Theta}(C)[n]) = m^2n^2(C,C). $$ By the adjunction formula (and the fact that the canonical bundle of $J(C)$ is trivial): $(C,C) = 2p_a(C)-2 = 2$.

Conclusion: When counted with multiplicity, $J_{\Theta}(C)[m]$ and $J_{\Theta}(C)[n]$ intersect in $2m^2n^2$ points.

The set $J(C)_{\Theta}[n]$ has the structure of a smooth irreducible algebraic curve, and the restriction of $J(C)\xrightarrow{\times n } J(C)$ to $C$ defines a morphism $J(C)_{\Theta}[n]\rightarrow C$ which is a finite etale cover with Galois group $J(C)[n]$. I don't think it is good to think of $J(C)_{\Theta}[n]$ as generalized torsion points. I think it's better to think of $J(C)_{\Theta}[n]$ as a curve which is an unramified cover of $C$. For example, while $J(C)[n]$ has $n^4$ points over $\bar{\mathbb{F}}_q$ (when $q$ is coprime to $n$), $J(C)_{\Theta}[n]$ has infinitely many $\bar{\mathbb{F}}_q$-points. Also by Riemann-Hurwitz the genus of $J(C)_{\Theta}[n]$ is $n^4+1$, so the Hasse-Weil bound tells you something about its $\mathbb{F}_q$-points.

For your second question about $J(C)_{\Theta}[m] \cap J(C)_{\Theta}[n]$, I will interpret this as a scheme theoretic intersection. (In other words, incorporating multiplicities.) We may calculate this using some facts about intersection theory on surfaces and line bundles on abelian varieties. As noted before, the curve $J(C)_{\Theta}[m]$ is the pullback of the divisor $C\subset J(C)$ (embedded using your implicitly chosen point $\mathcal{O}$). By the theorem of the square, we have the following linear equivalence of divisors on $J(C)$: $$J(C)_{\Theta}[n] \sim \frac{n^2+n}{2} C + \frac{n^2-n}{2} [-1]^*C.$$ Here $[-1]^*C$ is the image of $C$ under the $[-1]$ map. Since $[-1]^*C$ is algebraically equivalent to $C$ (even linearly equivalent if you choose $\mathcal{O}$ to be a Weierstrass point), we conclude that $$ (J_{\Theta}(C)[m],J_{\Theta}(C)[n]) = m^2n^2(C,C). $$ By the adjunction formula (and the fact that the canonical bundle of $J(C)$ is trivial): $(C,C) = 2p_a(C)-2 = 2$.

Conclusion: When counted with multiplicity, $J_{\Theta}(C)[m]$ and $J_{\Theta}(C)[n]$ intersect in $2m^2n^2$ points.

The set $J(C)_{\Theta}[n]$ has the structure of a smooth irreducible algebraic curve, and the restriction of $J(C)\xrightarrow{\times n } J(C)$ to $C$ defines a morphism $J(C)_{\Theta}[n]\rightarrow C$ which is a finite etale cover with Galois group $J(C)[n]$. I don't think it is good to think of $J(C)_{\Theta}[n]$ as generalized torsion points. I think it's better to think of $J(C)_{\Theta}[n]$ as a curve which is an unramified cover of $C$. For example, while $J(C)[n]$ has $n^2$ points over $\bar{\mathbb{F}}_q$ (when $q$ is coprime to $n$), $J(C)_{\Theta}[n]$ has infinitely many $\bar{\mathbb{F}}_q$-points. Also by Riemann-Hurwitz the genus of $J(C)_{\Theta}[n]$ is $n^2+1$, so the Hasse-Weil bound tells you something about its $\mathbb{F}_q$-points.

For your second question about $J(C)_{\Theta}[m] \cap J(C)_{\Theta}[m]$, I will interpret this as a scheme theoretic intersection. (In other words, incorporating multiplicities.) We may calculate this using some facts about intersection theory on surfaces and line bundles on abelian varieties. As noted before, the curve $J(C)_{\Theta}[m]$ is the pullback of the divisor $C\subset J(C)$ (embedded using your implicitly chosen point $\mathcal{O}$). By the theorem of the square, we have the following linear equivalence of divisors on $J(C)$: $$J(C)_{\Theta}[n] \sim \frac{n^2+n}{2} C + \frac{n^2-n}{2} [-1]^*C.$$ Here $[-1]^*C$ is the image of $C$ under the $[-1]$ map. Since $[-1]^*C$ is algebraically equivalent to $C$ (even linearly equivalent if you choose $\mathcal{O}$ to be a Weierstrass point), we conclude that $$ (J_{\Theta}(C)[m],J_{\Theta}(C)[n]) = m^2n^2(C,C). $$ By the adjunction formula (and the fact that the canonical bundle of $J(C)$ is trivial): $(C,C) = 2p_a(C)-2 = 2$.

Conclusion: When counted with multiplicity, $J_{\Theta}(C)[m]$ and $J_{\Theta}(C)[n]$ intersect in $2m^2n^2$ points.

The set $J(C)_{\Theta}[n]$ has the structure of a smooth irreducible algebraic curve, and the restriction of $J(C)\xrightarrow{\times n } J(C)$ to $C$ defines a morphism $J(C)_{\Theta}[n]\rightarrow C$ which is a finite etale cover with Galois group $J(C)[n]$. I don't think it is good to think of $J(C)_{\Theta}[n]$ as generalized torsion points. I think it's better to think of $J(C)_{\Theta}[n]$ as a curve which is an unramified cover of $C$. For example, while $J(C)[n]$ has $n^4$ points over $\bar{\mathbb{F}}_q$ (when $q$ is coprime to $n$), $J(C)_{\Theta}[n]$ has infinitely many $\bar{\mathbb{F}}_q$-points. Also by Riemann-Hurwitz the genus of $J(C)_{\Theta}[n]$ is $n^4+1$, so the Hasse-Weil bound tells you something about its $\mathbb{F}_q$-points.

For your second question about $J(C)_{\Theta}[m] \cap J(C)_{\Theta}[m]$, I will interpret this as a scheme theoretic intersection. (In other words, incorporating multiplicities.) We may calculate this using some facts about intersection theory on surfaces and line bundles on abelian varieties. As noted before, the curve $J(C)_{\Theta}[m]$ is the pullback of the divisor $C\subset J(C)$ (embedded using your implicitly chosen point $\mathcal{O}$). By the theorem of the square, we have the following linear equivalence of divisors on $J(C)$: $$J(C)_{\Theta}[n] \sim \frac{n^2+n}{2} C + \frac{n^2-n}{2} [-1]^*C.$$ Here $[-1]^*C$ is the image of $C$ under the $[-1]$ map. Since $[-1]^*C$ is algebraically equivalent to $C$ (even linearly equivalent if you choose $\mathcal{O}$ to be a Weierstrass point), we conclude that $$ (J_{\Theta}(C)[m],J_{\Theta}(C)[n]) = m^2n^2(C,C). $$ By the adjunction formula (and the fact that the canonical bundle of $J(C)$ is trivial): $(C,C) = 2p_a(C)-2 = 2$.

Conclusion: When counted with multiplicity, $J_{\Theta}(C)[m]$ and $J_{\Theta}(C)[n]$ intersect in $2m^2n^2$ points.