No. This condition implies that $t_i$, pulled back to $X$, vanishes only at $x_i$. Thus, $x_i$ is the entire fiber of the point where $t_i$ vanishes, so if $n$ is the degree of the map, then $\mathcal O(1) = \mathcal O(n x_i)$.

Thus $\mathcal O(nx_i)=\mathcal O(nx_j)$, so $x_i-x_j$ is an $n$-torsion divisor class, for all $i$, $j$.

To show this is impossible, it suffices to find two points $x_1$ and $x_2$ on a curve, such that $x_1-x_2$ is not a torsion divisor. To do this, let $E$ be an elliptic curve, let $x_1$ be the identity, and let $x_2$ be a non-torsion point. Then $x_1-x_2$ is not a torsion divisor class.