Let $I=\{\alpha\}$ and $k\in\mathbb{N}$. Consider the free group $F(I)$ constructed on $\{\alpha\}$. Let $\phi_\alpha$ be the canonical homomorphism of $\mathbb{Z}$ into $F(I)$. Let $r=\phi_\alpha(1)^k$ and let $N$ be the normal subgroup of $F(I)$ generated by $r$.
Let $f:F(I)\rightarrow F(I)/N$ be the canonical homomorphism and $e'$ the identity of $F(I)/N$. Write $x=f(\phi_\alpha(1))$. Then $x^k=e'$.
Why is $F(I)/N$ a finite group of order $k$? I understand that it is equal to the cyclic group $\{y\ |\ (\exists n)(n\in\mathbb{Z}\ \&\ y=x^n)\}$. I just don't get why it's finite: i.e. of order $k$?
The canonical homomorphism $\phi_{\alpha}$ is an isomorphism. Through this isomorphism the normal subgroup $N\subset F(I)$ corresponds to the normal subgroup $k\Bbb{Z}\subset\Bbb{Z}$, an the quotient $F(I)/N$ is canonically isomorphic to $\Bbb{Z}/k\Bbb{Z}$.
Closer to your exposition of the problem; because $F(I)$ is generated by $\alpha$ and the canonical homomorphism $f$ is surjective, its image $F(I)/N$ is generated by $f(\alpha)=f(\phi_{\alpha}(1))$. So $F(I)/N$ is cyclic, and from $x^k=e'$ it follows that its order divides $k$ if $k\neq0$.
Moreover, if $k\neq0$ then the order is precisely $k$ because $k$ is the least positive integer such that $x^n=e'$, because it is the least positive integer such that $\phi_{\alpha}(1)^n\in N$.
On the other hand, as you already note, if $k=0$ then $N=0$ and $F(I)/N\cong F(I)\cong\Bbb{Z}$, which is not finite.