Let $I=\{\alpha,\beta\}$ such that $\alpha\ne\beta$. Let $F(I)$ be the free group constructed on $I$ and $\phi_\alpha,\phi_\beta$ be the canonical injections of $\mathbb{Z}$ into $F(I)$. Write $r=\phi_\beta(1)^{-1}\phi_\alpha(1)^{-1}\phi_\beta(1)\phi_\alpha(1)$. Let $N_r$ be the normal subgroup of $F(I)$ generated by $r$.
Now, set $t_\alpha=(1,0)$ and $t_\beta=(0,1)$. Then there exists a unique surjective homomorphism $f_t:F(I)\rightarrow\mathbb{Z}\times\mathbb{Z}$ such that $f_t(\phi_\alpha(1))=t_\alpha$ and $f_t(\phi_\beta(1))=t_\beta$.
I want to prove that $\text{Ker}(f_t)=N_r$.
Let $w\in F(I)$ such that $f_t(w)=(0,0)$. Note that $w$ can be written uniquely in the form $\prod_{j=1}^n\phi_{i_j}(1)^{m(j)}$ for $n\in\mathbb{N}$, $m\in(\mathbb{Z}-\{0\})^{[1,n]}$, and $i\in\{\alpha,\beta\}^{[1,n]}$ such that $i_j\ne i_{j+1}$ for $1\leq j\leq n-1$. This implies that $$(0,0)=\sum_{j=1}^nm(j)\cdot f_t(\phi_{i_j}(1)).$$
I am now stuck because I am not sure sure how I can deduce that $\phi_{i_j}(1)\in\{r,r^{-1}\}$ for all $j\in[1,n]$.
Edit:
For $1\leq j\leq n$, either $m(j)\cdot f_t(\phi_{i_j}(1))=(m(j),0)$ or $m(j)\cdot f_t(\phi_{i_j}(1))=(0,m(j))$.