I think the flaw in your reasoning comes earlier in the proof. In the previous paragraph, Hatcher defines two moves that can be performed on a factorization of $[f]$. The second move is
Regard the term $[f_i]\in\pi_1(A_\alpha)$ as lying in the group $\pi_1(A_\beta)$ rather than $\pi_1(A_\alpha)$ if $f_i$ is a loop in $A_\alpha\cap A_\beta$.
Regarding this move, Hatcher asserts that
[This move] does not change the image of this element in the quotient group $Q=\ast_\alpha\, \pi_1(A_\alpha)/N$, by the definition of $N$
This is the step at which Hatcher is using the hypothesis that $N$ is normal. In particular, if $N$ were simply the subgroup generated by the elements $i_{\alpha\beta}(\omega)i_{\beta\alpha}(\omega)^{-1}$ (instead of the normal subgroup generated by these elements), this move would not necessarily preserve the image of this element in $G/N$.
We use the Seifert Van-Kampen Theorem to calculate the fundamental group of a connected graph. This is Hatcher Problem 1.2.5:
It is a fact in graph theory that any connected graph $X$ contains a maximal tree $M$, namely a contractible graph that contains all the vertices of $X$. Now if the maximal tree $M = X$, then we are done because for any $x_0 \in M$, $\pi_1(M,x_0) = \pi_1(X,x_0) = 0$ that is trivially free. Now suppose $M \neq X$. Then there is an edge $e_i$ of $X$ not in $M$. Observe that for each edge $e_i$ we get a loop going in $M \cup e_i$ about some point $x_0 \in M$. Now fix out basepoint $x_0$ to be in $M$ and suppose that the edges not in $M$ are $e_1,\ldots,e_n$. Then it is clear that
$$X = \bigcup_{i=1}^n \left(M \cup e_i\right).$$
The intersection of any two $M \cup e_i$ and $M \cup e_j$ contains at least $M$ and is path connected, so is the triple intersection of any 3 of these guys by the assumption that $X$ is a connected graph. So for any $x_0 \in M$, the Seifert-Van Kampen Theorem now tells us that
$$\pi_1(X,x_0) \cong \pi_1(M \cup e_1,x_0) \ast \ldots \ast \pi_1(M \cup e_n,x_0)/N$$
where $N$ is the subgroup generated by words of the form $l_{ij}(w)l_{ji}(w)^{-1}$, where $l_{ij}$ is the inclusion map from $\pi_1((M\cup e_i) \cap (M \cup e_j),x_0) = \pi_1(M \cup (e_i \cap e_j),x_0)$. Now observe that if $i \neq j$ then $M \cup (e_i \cap e_j) = M$ and since $\pi_1(M,x_0) = 0$ we conclude that any loop $w \in \pi_1(M \cup (e_i \cap e_j),x_0)$ in here is trivial. If $i = j$, $l_{ij}$ is just the identity so that our generators for $N$ are just
$$l_{ij}(w)l_{ji}(w)^{-1} = ww^{-1} = 1$$
completing our claim that $N$ was trivial. Now for each $i$, we have that $\pi_1(M\cup e_i,x_0)$ is generated by a loop that starts at $x_0$ and goes around the bounded complementary region formed by $M$ and $e_i$ and back to $x_0$ through the maximal tree. Such a path back to $M$ does not go through any other edge $e_j$ for $j$ different from $i$. It follows that $\pi_1(X,x_0)$ is a free group with basis elements consisting of loops about $x_0 \in M$ as described in the line before.
Best Answer
He means the smallest normal subgroup containing the elements. Typically, the subgroup they generate will not be normal. More generally, "the normal subgroup generated by..." usually means the smallest normal subgroup containing the elements, unless context makes it clear otherwise.