This is related to McCleary's User's guide to spectral sequences, Chpt 1, Example 1.H.
Suppose $V,W$ are graded $Q-$algebra and $E_2=V\otimes_QW$ is bigraded algebra converging to $H^\star=Q$. Assume $V\cong Q[x_{2n}]$ where $x_{2n}$ is of degree 2n to make $V$ commutative.
Consider differential action on $E_2^{0,\star}$. Take any $1\otimes u\in Q\otimes_QW^\star=E_2^{0,\star}$. $d(u)\in V\otimes_QW$ for sure. So $d(1\otimes u)=\sum_iv_i\otimes w_i$ for some $v_i,w_i$.
Suppose $u$ is of degree $l$ in $W^\star$. Then $(1\otimes u)\cdot\dots\cdot(1\otimes u) =(1\otimes u^k)$ where $\cdot$ is in $k$ times.
Now consider $d(1\otimes u^k)=d((1\otimes u)^k)$. In particular, this is asking for general derivation acting on an element $x^k$. Suppose $d:A\to A$ is degree 1 derivation with $x\in A$ with $A$ graded commutative. Consider $d(x^2)=dx(x)+(-1)^{|x|}xd(x)$. Now to commute $xd(x)$, I need to pick up an extra sign. In particular, I am not convinced that $d(1\otimes u^2)=2d(1\otimes u)(1\otimes u)$.
The book says $d(1\otimes u^k)=k\sum_iv_i\otimes w_iu^{k-1}$. If I specialize to $k=2$, I have $d(1\otimes u^2)=2d(1\otimes u)(1\otimes u)$
$\textbf{Q:}$ How did the book reach conclusion that $d(1\otimes u^k)=kd(1\otimes u)(1\otimes u^{k-1})$? If $u$ is of even degree, I could believe this is the case but this may not be the case. The goal is to check $W=\wedge[y_{2n+1}]$.(i.e. $W$ is exterior algebra generated by 1 variable.)
$\textbf{Q':}$ Given $V=Q[x_{2n}],W=\wedge[y_{2n+1}]$ graded algebra as above, I can construct $E_2$. However, it seems that $E_2$'s differential map is extra information from $V,W$ graded algebra structure.
Regarding Q, I agree that this may not be the case if $u$ is in odd degree. In his proof, though, if $u$ is in odd degree, then $u^2$ will be forced to be zero ("If $W^*$ contains any other elements ..." on p. 21). So the formula is only used when $u$ is in even degree.
Regarding Q', it's not a question but a statement. In general, the differential $d_r$ of a spectral sequence is extra information beyond the identification of $E_r$. Sometimes you can deduce what $d_r$ must be, as in this example, but certainly not always.