Let me add a quick proof, also working in characteristic zero, that doesn't rely on classification (this proof is essentially originally due to S'andor, see his Duke paper on rational singularities). It does rely on canonical modules and a certain functoriality for them. I will use the following characterization of rational singularities though (which I believe is originally due to Kempf), $Z$ has rational singularities if and only if it is Cohen-Macaulay and for a resolution $\pi : Z' \to Z$, we have $$\pi_* \omega_Z' \cong \omega_Z.$$

Suppose that $R \subseteq S$ is the invariant subring of some finite group action. We know $f : R \to S$ splits as a map of $R$-module (trace map splits everything).

Take a resolution $\pi : X \to \text{Spec} R$ and another proper birational map $\pi' : Y \to \text{Spec} S$ such that $f \circ \pi'$ factors through $\pi$ (we can always do this, we can even assume $Y$ is also smooth if we feel like it). So $f \circ \pi' = \pi \circ \eta$ and $\eta : Y \to X$ is some proper map.

Now, we have the following map

$$\pi_* \eta_* \omega_Y \to \omega_R$$

which can either be factored as $$\pi_* \eta_* \omega_Y \to \pi_* \omega_X \to \omega_R$$ or $$f_* \pi'_* \omega_Y \to f_* \omega_S \to \omega_R.$$

In the second factorization, the first map is surjective because $S$ is regular and the second map $f_* \omega_S \to \omega_R$ is surjective because $R \to S$ splits (that map can be obtained by applying $Hom_R( \cdot , \omega_R)$ for instance). Thus the original map is surjective and so is $\pi_* \omega_X \to \omega_R$.

If we want to prove rational singularities, we now just need to prove Cohen-Macaulayness. But that is obvious for normal surfaces. Of course, the Cohen-Macaulayness of a summand of a regular ring is also pretty easy ($H^i_{\mathfrak{m}}(R) \to H^i_{\mathfrak{m}}(S)$ injects).

Let me add a quick proof, also working in characteristic zero, that doesn't rely on classification (this proof is essentially originally due to S'andor, see his Duke paper on rational singularities). It does rely on canonical modules and a certain functoriality for them. I will use the following characterization of rational singularities though (which I believe is originally due to Kempf), $Z$ has rational singularities if and only if it is Cohen-Macaulay and for a resolution $\pi : Z' \to Z$, we have $$\pi_* \omega_{Z'} \cong \omega_Z.$$

Suppose that $R \subseteq S$ is the invariant subring of some finite group action. We know $f : R \to S$ splits as a map of $R$-module (trace map splits everything).

Take a resolution $\pi : X \to \text{Spec} R$ and another proper birational map $\pi' : Y \to \text{Spec} S$ such that $f \circ \pi'$ factors through $\pi$ (we can always do this, we can even assume $Y$ is also smooth if we feel like it). So $f \circ \pi' = \pi \circ \eta$ and $\eta : Y \to X$ is some proper map.

Now, we have the following map

$$\pi_* \eta_* \omega_Y \to \omega_R$$

which can either be factored as $$\pi_* \eta_* \omega_Y \to \pi_* \omega_X \to \omega_R$$ or $$f_* \pi'_* \omega_Y \to f_* \omega_S \to \omega_R.$$

In the second factorization, the first map is surjective because $S$ is regular and the second map $f_* \omega_S \to \omega_R$ is surjective because $R \to S$ splits (that map can be obtained by applying $Hom_R( \cdot , \omega_R)$ for instance). Thus the original map is surjective and so is $\pi_* \omega_X \to \omega_R$.

If we want to prove rational singularities, we now just need to prove Cohen-Macaulayness. But that is obvious for normal surfaces. Of course, the Cohen-Macaulayness of a summand of a regular ring is also pretty easy ($H^i_{\mathfrak{m}}(R) \to H^i_{\mathfrak{m}}(S)$ injects).

Let me add a quick proof, also working in characteristic zero, that doesn't rely on classification (this proof is essentially originally due to S'andor, see his Duke paper on rational singularities). It does rely on canonical modules and a certain functoriality for them. I will use the following characterization of rational singularities though (which I believe is originally due to Kempf), $Z$ has rational singularities if and only if it is Cohen-Macaulay and for a resolution $\pi : Z' \to Z$, we have $$\pi_* \omega_Z' \cong \omega_Z.$$

Suppose that $R \subseteq S$ is the invariant subring of some finite group action. We know $f : R \to S$ splits as a map of $R$-module (trace map splits everything).

Take a resolution $\pi : X \to \text{Spec} R$ and another proper birational map $\pi' : Y \to \text{Spec} S$ such that $\pi' = \pi \circ \eta$ factors through $\pi$ (we can always do this, we can even assume $Y$ is also smooth if we feel like it), so $\eta : Y \to X$ is some proper map.

Now, we have the following map

$$\pi_* \eta_* \omega_Y \to \omega_R$$

which can either be factored as $$\pi_* \eta_* \omega_Y \to \pi_* \omega_X \to \omega_R$$ or $$f_* \pi'_* \omega_Y \to f_* \omega_S \to \omega_R.$$

In the second factorization, the first map is surjective because $S$ is regular and the second map $f_* \omega_S \to \omega_R$ is surjective because $R \to S$ splits (that map can be obtained by applying $Hom_R( \cdot , \omega_R)$ for instance). Thus the original map is surjective and so is $\pi_* \omega_X \to \omega_R$.

If we want to prove rational singularities, we now just need to prove Cohen-Macaulayness. But that is obvious for normal surfaces. Of course, the Cohen-Macaulayness of a summand of a regular ring is also pretty easy ($H^i_{\mathfrak{m}}(R) \to H^i_{\mathfrak{m}}(S)$ injects).

Let me add a quick proof, also working in characteristic zero, that doesn't rely on classification (this proof is essentially originally due to S'andor, see his Duke paper on rational singularities). It does rely on canonical modules and a certain functoriality for them. I will use the following characterization of rational singularities though (which I believe is originally due to Kempf), $Z$ has rational singularities if and only if it is Cohen-Macaulay and for a resolution $\pi : Z' \to Z$, we have $$\pi_* \omega_Z' \cong \omega_Z.$$

Suppose that $R \subseteq S$ is the invariant subring of some finite group action. We know $f : R \to S$ splits as a map of $R$-module (trace map splits everything).

Take a resolution $\pi : X \to \text{Spec} R$ and another proper birational map $\pi' : Y \to \text{Spec} S$ such that $f \circ \pi'$ factors through $\pi$ (we can always do this, we can even assume $Y$ is also smooth if we feel like it). So $f \circ \pi' = \pi \circ \eta$ and $\eta : Y \to X$ is some proper map.

Now, we have the following map

$$\pi_* \eta_* \omega_Y \to \omega_R$$

which can either be factored as $$\pi_* \eta_* \omega_Y \to \pi_* \omega_X \to \omega_R$$ or $$f_* \pi'_* \omega_Y \to f_* \omega_S \to \omega_R.$$

In the second factorization, the first map is surjective because $S$ is regular and the second map $f_* \omega_S \to \omega_R$ is surjective because $R \to S$ splits (that map can be obtained by applying $Hom_R( \cdot , \omega_R)$ for instance). Thus the original map is surjective and so is $\pi_* \omega_X \to \omega_R$.

If we want to prove rational singularities, we now just need to prove Cohen-Macaulayness. But that is obvious for normal surfaces. Of course, the Cohen-Macaulayness of a summand of a regular ring is also pretty easy ($H^i_{\mathfrak{m}}(R) \to H^i_{\mathfrak{m}}(S)$ injects).