In this answer all loops are based at the identity unless specified otherwise.
Recall that by the homotopy lifting property, since $p: G \rightarrow G/H$ is a covering map, any loop $\gamma: [0,1] \rightarrow G/H$ has a unique lift $\tilde \gamma: [0,1] \rightarrow G$ such that $\tilde \gamma(0) = 1_G$. Consider the map $f: \pi_1(G/H) \rightarrow H$ given by $f(\gamma) = \tilde\gamma(1)$. This is well-defined because the lift is unique, and by the homotopy lifting property again any homotopy $\gamma \simeq \gamma'$ lifts to a homotopy of paths in $X$; since $p(\gamma_t(1))=e$, and $H$ is discrete, we see that $\tilde\gamma'(1)=\tilde\gamma_t(1)=\tilde\gamma(1)$ for all $t$. The codomain is $H$ because $\gamma(1) = 1_{G/H}$, $\tilde \gamma(1) \in H$.
1) This is a homomorphism. For two homotopy classes of loops with representatives $\gamma, \eta$, if $f(\gamma) = h_1$ and $f(\eta) = h_2$, then we may lift $\eta$ to $\tilde\eta_{h_1}(t) = h_1\tilde \eta(t)$, whose basepoint is $h_1$ rather than $1_G$; clearly \tilde\eta_{h_1}(1)=h_1h_2. Since lifts are unique, the lift of the concatenation $\gamma*\eta$ first follows $\tilde \gamma$ and then $\tilde \eta_{h_1}$, so $\widetilde{\gamma * \eta}(1) = h_1h_2$, so it follows that $f$ is a homomorphism.
2) This map is injective. To see this, pick a loop $\gamma$ such that $\tilde \gamma(1) = 1_G$. Since $G$ is simply connected, $\tilde \gamma$ is null-homotopic (keeping the basepoint fixed, as usual). Projecting this homotopy to $G/H$ shows that $\gamma$ is null-homotopic as well. So the kernel of $f$ is trivial.
3) This map is surjective. Indeed, since $G$ is path-connected, fix $h \in H$ and pick a path $\gamma: I \rightarrow G$ with $\gamma(0) = 1_G$, $\gamma(1) = h$. This projects to a loop in $G/H$; its class $[\gamma] \in \pi_1(G/H)$ has $f([\gamma]) = h$.
So $f$ is an isomorphism, and $\pi_1(G/H) \cong H$.
