Is it possible to find a group $G_0$ and a proper subgroup $H$ such that for all morphism $f,g$ from $G_0$ to $G_1$ such that $f_{\scriptscriptstyle{\vert H}}=g_{\scriptscriptstyle{\vert H}}$ implies $f=g$ ?
It seems to be a natural question, but I can't find any example, of proof that is not possible. Any ideas ?
No. Your question is equivalent to asking whether it's possible for the inclusion $f : H \to G$ of a proper subgroup into a group $G$ to be an epimorphism, and it's not: the epimorphisms in groups are precisely the surjective homomorphisms. (This isn't a purely formal statement, and isn't true in related cases such as for rings.)
The proof uses the pushout or amalgamated free product $G \ast_H G$. This construction is natural in the context of this problem because it is the universal group equipped with a pair of maps $f_0, f_1 : G \to G \ast_H G$ (namely the inclusions of the first and second factors) which agree on $H$. Another name for it is the cokernel pair of $f$.
From here it suffices to show that if $f_0 = f_1$ then $H = G$. But it's known that if $f_0(g) = f_1(g) \in G \ast_H G$ then the common value $f_0(g) = f_1(g)$ must lie in $H$. In other words, the two factors in a pushout of groups interact as little as possible.