If the map $f^* \colon H^0(Y,\mathcal{L}) \to H^0(X,f^*\mathcal{L})$ is surjective then there is a commutative diagram $$ \require{AMScd} \begin{CD} X @>>> \mathbb{P}(H^0(X,f^*\mathcal{L})^\vee) \\ @VfVV @VVV \\ Y @>>> \mathbb{P}(H^0(Y,\mathcal{L})^\vee), \end{CD} $$ where the right vertical arrow is an embedding and the bottom arrow is a priori rational. But since $f$ is proper and dominant, it is surjective, hence the bottom arrow is regular. Moreover, it follows that $f$ is a closed embedding. But then $f$ is an isomorphism.

If the map $f^* \colon H^0(Y,\mathcal{L}) \to H^0(X,f^*\mathcal{L})$ is surjective then there is a commutative diagram $$ \require{AMScd} \begin{CD} X @>>> \mathbb{P}(H^0(X,f^*\mathcal{L})^\vee) \\ @VfVV @VVV \\ Y @>>> \mathbb{P}(H^0(Y,\mathcal{L})^\vee), \end{CD} $$ where the right vertical arrow is an embedding and the bottom arrow is a priori rational. But since $f$ is proper and dominant, it is surjective, hence the bottom arrow is regular. Moreover, it follows that $f$ is a closed embedding. But then $f$ is an isomorphism.

EDIT. If you want to check the surjectivity of $H^0(Y,\mathcal{L}) \otimes \mathcal{O}_Y \to \mathcal{L}$ at point $y \in Y$, choose $x \in X$ such that $y = f(x)$, and consider the commutative diagram $$ \require{AMScd} \begin{CD} H^0(Y,\mathcal{L}) @>>> \mathcal{L}_y \\ @V f^* VV @| \\ H^0(X,f^*\mathcal{L}) @>>> (f^*\mathcal{L})_x. \end{CD} $$ The left vertical arrow is surjective by assumption, and the bottom arrow is surjective, because $f^*\mathcal{L}$ is very ample. Therefore, the top arrow is surjective as well.