We have that the map $\Pi^{pq}_{\gamma}: T_pS \rightarrow T_qS$ that takes $v_0 \in T_pS$ to $v_1 \in T_qS$ is called parallel transport from $p$ to $q$ along $\gamma$.
$\Pi^{pq}_{\gamma}: T_pS \rightarrow T_qS$ is a linear map and an isometry.
How can we show that the parallel transport map $\Pi^{pq}_{\gamma}: T_pS \rightarrow T_qS$ is invertible?
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The inverse is given by parallel transport along the inverse path.
Let $\gamma : [0, 1] \to S$, with $\gamma(0) = p$ and $\gamma(1) = q$. If $v_0 \in T_pS$ and $X$ is a vector field on the image of $\gamma$ such that $X_p = v_0$ and $\nabla_{\gamma'}X = 0$, then we set $\Pi^{pq}_{\gamma}(v_0) = X_q$ (which you called $v_1$).
Now let $\hat{\gamma}(t) = \gamma(1-t)$, so $\hat{\gamma} : [0, 1] \to S$, $\hat{\gamma}(0) = \gamma(1) = q$ and $\hat{\gamma}(1) = \gamma(0) = p$. Note that $\hat{\gamma}' = -\gamma'$ so if $X$ is the vector field above, $\nabla_{\hat{\gamma}'}X = \nabla_{-\gamma'}X = -\nabla_{\gamma'}X = 0$ and $X_q = v_1$, so $\Pi^{qp}_{\hat{\gamma}}(v_1) = X_p = v_0$.
Note, the same argument works for parallel transport on any vector bundle with connection. The only thing that changes is that $X$ is replaced by a section of the bundle which is covariantly constant along $\gamma$.