I am not sure what the appropriate way to map a relation to another is (the relations do not actually exist in the group, but rather in the free group on the generators that one is quotienting by in order to obtain $G$), but for surjectivity: it is sufficient for the homomorphism to map some generating set to some other generating set, and it is also necessary for it to map every generating set to another generating set. To see this, consider the following
Lemma. Let $f:G\to H$ be a homomorphism, and $X\subseteq G$ a subset. Then the image of the span of the set $X$ under the map $f$ is the span of the image of $X$ under $f$; i.e. $f^*\langle X\rangle=\langle f^*X\rangle$.
Proof. (formal enough for my personal satisfaction anyway)
$\begin{array}{cl} f^*\langle X\rangle & =f^*\{x_1^{e_1}\cdots x_m^{e_m}:x_j\in X\} \\ & =\{f(x_1)^{e_1}\cdots f(x_m)^{e_m}:x_j\in X\} \\ & = \{u_1^{e_1}\cdots u_m^{e_m}:u_j\in f^*X\} \\ & =\langle f^*X\rangle.\end{array}$
And hence
Corollary. Suppose $\langle X\rangle=G$, i.e. $X$ is a generating set. Then $f:G\to H$ is surjective iff $f^*G=H$ iff $f^*\langle X\rangle=H$ iff $\langle f^*X\rangle=H$ iff $f^*X$ is a generating set for $H$.