The answer to your second question is no. If $x\notin J(\mathbb Q)$, then the condition $\pi(x)\in X(\mathbb Q)$ implies that $\mathbb Q(x)/\mathbb Q$ is a quadratic extension. If you take a point $y$ with the same properties, then it is highly unlikely that $\mathbb Q(x)\mathbb Q(y)$, and then $\mathbb Q(x+y)$ is most likely a quartic extension of $\mathbb Q$, so $\pi(x+y)$ will not be in $X(\mathbb Q)$. You might get some insight if you consider the analogous question for an elliptic curve and the map $E\to E/\{\pm1\}\cong\mathbb P^1$.

The answer to your second question is no. If $x\notin J(\mathbb Q)$, then the condition $\pi(x)\in X(\mathbb Q)$ implies that $\mathbb Q(x)/\mathbb Q$ is a quadratic extension. If you take a point $y$ with the same properties, then it is highly unlikely that $\mathbb Q(x)\cong\mathbb Q(y)$, and then $\mathbb Q(x+y)$ is most likely a quartic extension of $\mathbb Q$, so $\pi(x+y)$ will not be in $X(\mathbb Q)$. You might get some insight if you consider the analogous question for an elliptic curve and the map $E\to E/\{\pm1\}\cong\mathbb P^1$.