$N$ is the normal subgroup of $*_\alpha \pi_1(A_\alpha)$ generated by all words of the form $i_{\alpha\beta}(\omega)i_{\beta\alpha}(\omega)^{-1}$ with $\omega \in \pi_1(A_\alpha \cap A_\beta)$.
Hatcher does not formulate the second move very precisely. He "really" means that if $g_j$ is a loop in $A_\alpha \cap A_\beta$, i.e. $ [g_j] \in \pi_1(A_\alpha \cap A_\beta)$, then we are allowed to replace $[f_j] = i_{\alpha\beta}([g_j]) \in \pi_1(A_\alpha)$ by $[f'_j] = i_{\beta\alpha}([g_j]) \in \pi_1(A_\beta)$.
To understand this, note that by definition the word $[f_j][f'_j]^{-1}$ is contained in $N$. Similarly, since $[g_j]^{-1} \in \pi_1(A_\alpha \cap A_\beta)$, we see that also $[f_j]^{-1}[f'_j] = i_{\alpha\beta}([g_j])^{-1}i_{\beta\alpha}([g_j]) = $ $i_{\alpha\beta}([g_j]^{-1})i_{\beta\alpha}([g_j]^{-1})^{-1} \in N$. This implies that if $w$ is an arbitrary word containing $[f_j]^\epsilon$, $\epsilon = \pm 1$, and the word $w'$ is obtained by replacing $[f_j]^\epsilon$ with $[f'_j]^\epsilon$, then $w,w'$ have the same equivalence class in $Q$. To see this, note that we can write $w = u [f_j]^\epsilon v$ and $w' = u [f'_j]^\epsilon v$ with suitable words $u, v$ in $*_\alpha \pi_1(A_\alpha)$. Then $w(w')^{-1} = u [f'_j]^\epsilon v v^{-1} [f'_j]^{-\epsilon}u^{-1} = u [f'_j]^\epsilon [f'_j]^{-\epsilon}u^{-1} \in N$ because $[f'_j]^\epsilon [f'_j]^{-\epsilon} \in N$ and $N$ is a normal subgroup of $*_\alpha \pi_1(A_\alpha)$.
In your example you have $[g]\in \pi_1(A_1\cap A_3)$, $[f_1] = i_{13}([g]) \in \pi_1(A_1)$, $[f_2]\in \pi_1(A_2)$. With $[f_3] = i_{31}([g]) \in \pi_1(A_3)$ you see that $[f_1][f_2]$ and $[f_3][f_2]$ are equivalent, but it is not true that$[f_1][f_2]$ and $[f_2][f_1]$ are equivalent.
Edited:
Due to the discussion in the comments let me "go back to the roots".
A map $f : X \to Y$ is determined by its domain $X$, its range $Y$ and the rule of assignment $X \ni x \mapsto f(x) \in Y$. The range is an essential constituent of $f$. If you have a space $Y' \supset Y$, you may regard $f$ as map into $Y'$, but the precise interpretation is that you replace $f$ by the map $i \circ f : X \to Y'$ (where $i : Y \to Y'$ denotes the inclusion map). The maps $f$ and $i \circ f$ are not the same because the have different ranges (although you have $f(x) = (i \circ f)(x)$ for all $x \in X$). Conversely, if you have a map $f' : X \to Y'$ such that $f(X) \subset Y$, you may regard $f$ as map into $Y$, but the precise interpretation is that there exists a unique map $f : X \to Y$ such that $f' = i \circ f$.
Precision is no nitpicking. This is of particular importance if you consider homotopy classes of maps. The range determines how much space is available for deformations. Two non-homotopic maps $f, g : X \to Y$ may become homotopic in a bigger $Y' \supset Y$ which means that $i \circ f, i \circ g$ are homotopic. As an example take $id : S^1 \to S^1$ which is not null-homotopic, but becomes null-homotopic in $\mathbb R^2$.
Hatcher does not address this point, probably it seems self-evident to him. When he says that $g_{k-1} \circ f \circ g_k$ is a loop in $A_k$, he means that the map $g_{k-1} \circ f \circ g_k : I \to X$ has the property $(g_{k-1} \circ f \circ g_k)(I) \subset A_k$ and thus can be written as $g_{k-1} \circ f \circ g_k = \iota_k \circ \phi_k$ with $\phi_k : I \to A_k$, where $\iota_k : A_k \to X$ denotes the inclusion map.
Then $[\phi_k] \in \pi_1(A_k)$ and you get $i_k([\phi_k]) = (\iota_k)_*([\phi_k]) = [\iota_k \circ \phi_k] = [g_{k-1} \circ f \circ g_k] \in \pi_1(X)$.
