Note that exactness depends only on the underlying objects, and not on the differentials. It follows that if each sequence at coordinate is exact, then the sequence of total complexes is too, because the direct sum of exact sequences is exact. On the other hand, the morphism you have defined is not how the map $\mathrm{Tot}(f)$ for a map $f:C\to D$ of double complexes works. Rather, an element in $\mathrm{Tot}(C)$ of the form $(\ldots,c_{0,n},c_{1,n-1},\ldots)$ gets sent to $(\ldots,fc_{0,n},fc_{1,n-1},\ldots)$.
For completeness, I think I can cover the case where $A_{\bullet}$ and $B_{\bullet}$ are exact myself now. I will try to stick to the notations used in the other posts.
If $c \in C_n$ with $d(c)=0$, there is, by the surjectivity of $\beta_n$, a $b$ in $B_n$ such that $$\beta_n(b)=c.$$
Then $$\beta_{n-1}(d (b)) = d(\beta_n (b)) = d(c) = 0$$ and thus, by exactness of the n-1-th row, there is an $a$ in $A_{n-1}$ with $\alpha_{n-1}(a) = d(b)$.
For $a$ we have $$\alpha_{n-2}(d(a)) = d(\alpha_{n-1}(a)) = d(d(b)) = 0$$ and, since $\alpha_{n-2}$ is injective, $d(a)=0$ follows.
Then, by exactness of $A_{\bullet}$ we have an $a'$ in $A_n$, such that $d(a')=a$.
Consider $b-\alpha_n(a')$ in $B_n$. We have $$d(b - \alpha_n(a'))=d(b) - d(\alpha_n(a')) = d(b) - \alpha_{n-1}(d(a')) = d(b) - d(b) = 0,$$ thus by exactness of $B_{\bullet}$, there is a $b'$ in $B_{n+1}$, such that $d(b')=b-\alpha_n(a')$.
Finally, $\beta_{n+1}(b')$ is the desired pre-image of $c$ in $C_{n}$, since $$d(\beta_{n+1}(b'))=\beta_n(d(b'))=\beta_n(b) - \beta_n(\alpha_n(a')) = c.$$
Thanks, everyone!
Best Answer
Weibel's construction is perhaps a bit obscure (although effective if you mean to prove that such resolution exists in a quick way). The idea of the horseshoe lemma, among other things, is to obtain a split resolution of a given short exact sequence of modules. Thus, you want to define an appropriate differential $d : P'\oplus P'' \longrightarrow P'\oplus P''$ that make the desired diagram commute, where the resolution of your exact sequence is the (canonically) split exact sequence of complexes
$$ P' \longrightarrow P'\oplus P''\longrightarrow P''$$
You don't need to check the middle complex is a resolution once you obtained one, because it is automatically exact by the homology LES of the last short exact sequence.
If we write $\varepsilon',\varepsilon''$ for the augmentations for $A'$ and $A''$, then writing $\varepsilon_0'+\varepsilon_0''$ for the augmentation of $A$ where $$\varepsilon_0':P_0' \longrightarrow A$$ $$\varepsilon_0'':P_0'' \longrightarrow A$$
following the arrows gives
$$\varepsilon_0' =f\varepsilon'$$ $$\varepsilon ''\pi = g\varepsilon_0''$$
and because $P_0''$ is projective you can solve for $\varepsilon_0''$. Now $d$ is a matrix of morphisms $\begin{pmatrix} f_1 & f_2 \\ g_1&g_2 \end{pmatrix}$. In particular, if we want commutativity at $P'$ we need that (follow the arrows, again)
$$f_1x'= d'x'$$ $$g_1x'=0$$
and commutativity at $P''$ gives that $g_2x''=d''x''$. We have shown that $$d=\begin{pmatrix} d' & f_2 \\ 0&d'' \end{pmatrix}$$ which is what Weibel wanted.
Thus it suffices to determine $f_2:P''\longrightarrow P'$ inductively so that $d^2=0$, or, what is the same, so that $$d'f_2 +f_2 d''=0$$ and I leave that to you. Note we already determined the map at degree $0$, where it looks a bit different. At any rate, the proof is an inductive lifting argument.