I have been reading the appendix in Serre's Local fields, to do with explicit computations of cup products (pg 176), but I'm stuck on one bit of lemma 4. It goes as follows

Let B be a $G$-module, $u: G \times G \rightarrow B$ a 2-cocycle and $\bar{u} \in H^2(G,B)$. Then for all $s \in G$ with $\bar{s} \in \widehat{H}^{-2}(G,\mathbb{Z})$. Then $$\bar{s} \cup \bar{u} = \bar{a}, \qquad \text{with } a=\sum_{t \in G} u(t,s)$$

Now to prove it he begins by taking a sequence $0 \rightarrow B \rightarrow B' \rightarrow B'' \rightarrow 0$ with $B'$ induced. Now since $H^2(G,B')=0$, there is a 1-cochain $f': G \rightarrow B$ such that $$u(x,y)=xf'(y)-f'(xy) + f(x), \qquad x,y \in G.$$ Then composing $f'$ with $B' \rightarrow B''$ we get a 1-cocycle $f'':G \rightarrow B$, with $d(\bar{f''})=\bar{u}$. From which he deduces that $$\bar{s} \cup \bar{u} = \bar{s} \cup d(\bar{f''})=d(\bar{s} \cup \bar{f''})$$

Now by the previous lemma we know $\bar{s} \cup \bar{f''}= \bar{f''(s)}_0$ (here the subscript 0 denotes its class in $H^{-1}(G,B''))$. Now this is where I'm stuck he then says that $d(\bar{s} \cup \bar{f''})=d( \overline{f''(s)}_0)= \overline{N(f'(s))}^0$ (where the superscript 0 denotes its class in $\widehat{H}^0$), my question is why does the $f''$ change to $f'$? I cant see why this is so, he also does something similar in lemma 2 of this same appendix, but this is the lemma I need to use.

Thank you