What exactly is the isomorphism ("via" $p_*$) with $G \cong \pi_1 (\tilde{X}_G, \tilde{x}_0)$?
It cannot be $p_*$ restricted to its image, i.e. $\tilde{p}_* : \pi_1 (\tilde{X}_G, \tilde{x}_0) \rightarrow G$. This map is not injective in general.
And does this isomorphism always hold for any subgroup G of a fundamental group?
The isomorphism is indeed $p_*$, because in the context of Theorems 10.34 and 10.46 the homomorphism $p_*$ is injective.
Presumably the textbook has already proved that for any covering map $p : \tilde X \to X$ and any $\tilde x \in \tilde X$ and $x=p(\tilde x) \in X$, the induced homomorphism $p_* : \pi_1(\tilde X,\tilde x) \to \pi_1(X,x)$ is injective. The proof usually comes rather early in the theory of covering maps, because it is a straightforward application of the homotopy lifting lemma. Here's a brief sketch. For any closed path $\gamma : [0,1] \to \tilde X$ based at $\tilde x$, the composition $p \circ \gamma : [0,1] \to X$ is a closed path based at $x$. If there exists a path homotopy from $p \circ \gamma$ to the constant path at $x$ then, by applying the homotopy lifting lemma, one obtains a path homotopy from $\gamma$ to the constant path at $\tilde x$. In other words, if $p_*[\gamma]$ is the identity then $[\gamma]$ is the identity. This implies that $p_*$ is injective.
Combined with the conclusion of Theorem 10.34, it follows that $p_* : \pi_1(\tilde X_G,\tilde x_0) \to G$ is an isomorphism.
