I am trying to understand a claim on page 162 of Bott and Tu's "Differential forms in Algebraic Topology", for which I'll recall some notation briefly. Say $K = \oplus_{p,q \geq 0} K^{p,q}$ is a double complex with two differentials $\delta: K^{p,q} \rightarrow K^{p+1, q}$ and $d:K^{p,q} \rightarrow K^{p,q+1}$. Look at $K_{p,q} := \oplus_{i \geq p} K^{i,q}$ (this is my own notation). We consider the double complex $\oplus_{p, q \geq 0} K_{p,q}$ and take its diagonal single complex $A := \oplus_{k \geq 0} A^k$ where $A^k:= \oplus_{p+q = k} \left( K_{p,q} \right)$.
We consider also $K_p := \oplus_{q \geq 0} \oplus_{i \geq p} K^{i,p}$ and the quotient $K_p/K_{p+1}$ as a double complex and do the same procedure to obtain a single complex $B = \oplus_{k \geq 0} B^k$, where $B^k = \oplus_{p+q = k} K_{p,q}/K_{p+1, q} = \oplus_{p+q=k} K^{p,q}$ (I hope I'm not mistaken about this).
Pictorially $A^k$, $B^k$ and $K_p$ look like (for $p=2, k=4$):
Now we have a differential $D:A \rightarrow A$ which defined like $D:A^k \rightarrow A^{k+1}$, $D = \delta + (-1)^p d$ where $p$ is the first degree of the argument. It is then said that this differential restricts to $D = (-1)^p d:B \rightarrow B$ in order to make the map $j: A \rightarrow B$ into a chain complex morphism, where $j:A^k \rightarrow B^k$ just takes a set of elements in the cells of $A^k$ and only keeps those elements on the diagonal which is $B^k$. At least, this is the only way I can see it defined, please correct me if I'm wrong.
My first question is, what am I getting wrong in my understanding of the situation which makes it that my $j$ is not a chain complex morphism?
What I am actually trying to understand is why can we represent an element of $E_2$ as is claimed in 162, as the class of a $b$ with $db = 0$ and $\delta b$ being $d$-exact. My attempt to see this (which fails) is:
Firstly, $E_2$ is the cohomology of $E_1$ w.r.t. some induced maps $j_1$ and $k_1$, and $E_1$ is by definition $H_D(B)$, so $H_d(B)$ (because $D = (-1)^p d$) i.e. simply the cohomology we get by going vertically with $d$ - so an element in $H^k_d(B)$ is just a sequence of cohomology classes $[b_{0,k}], [b_{1, k-1}], \ldots, [b_{k, 0}]$ all satisfying $d b_{i,j} = 0$.
Now the $k_1$ is the map $\delta$ seen as $k_1: H_d(B) \rightarrow H_D(A)$, so going from $H_d^k(B)$ to $H_D^k(A)$ i.e. taking a given sequence $[b_{0,k}], [b_{1, k-1}], \ldots, [b_{k, 0}]$ as above and mapping them to $[j_1 k_1 ([b_{i,j}])$. So what this simply does is, it takes a collection of elements on the diagonal figured in red in the picture and shifts them via $\delta$ to the next diagonal ($B^{k+1}$). So the condition that the sequence $[b_{0,k}], [b_{1, k-1}], \ldots, [b_{k, 0}]$ is in $\ker j_1 k_1$ simply means that all $b_{i,j}$ are $\delta$-closed.
In conclusion, I obtain (wrongly) that elements of $E_2$ must simply be elements in $B$ which are $d$ and $\delta$ closed.
Please help me understand what I am getting wrong.