$\newcommand\CC{\mathbb{C}}$Since $a_1$, $\dots$, $a_n$ are different, the map $\psi:\CC[t^{\pm1}]/(f)\to\CC^n$ (the codomain being the direct product of $n$ copies of $\CC$) given by $\psi(\overline g)=(g(a_1),\dots,g(a_n))$ is an isomorphism of rings. The image of the composition $\CC[t^{\pm2}]\to\CC[t^{\pm1}]\to\CC[t^{\pm1}]/(f)\xrightarrow{\;\psi\;}\;\CC^n$ contains the $n$ vectors $(1,\dots,1), \qquad (a_1^2,\dots,a_n^2), \qquad (a_1^4,\dots,a_n^4), \qquad\dots,\qquad (a_1^{2(n-1)},\dots,a_n^{2(n-1)}),$ because they are in fact the elements $\psi(\overline1)$, $\psi(\overline{t^2})$, $\psi(\overline{t^4})$, $\dots$, $\psi(\overline{t^{2(n-1)}})$. Since the numbers $a_1^2$, $\dots$, $a_n^2$ are different, these $n$ vectors are linearly independent: this follows from the usual Vandermonde argument. Therefore, the image of that composition is surjective.
This implies immediately that your map is surjective.