$\require{AMScd}$The following is taken from these notes by Daniel Murfet.
Let $ \cdots \subseteq F^{p + 1}(C) \subseteq F^p(C) \subseteq F^{p - 1}(C) \subseteq \cdots$ be a filtration of a complex $C$ in an abelian category.
There is either a mistake or I don't understand something. I think that $\ddot{A^{pq}_r} \subseteq \ddot{A^{pq}_{r + 1}}$. Indeed, $A^{pq}_r$ defined by the following pullback
$$\begin{CD} A^{pq}_r @>>> F^p(C^{p + q}) \\ @VVV @VVV \\ F^{p + r}(C^{p + q + 1}) @>>> F^p(C^{p + q + 1}) \end{CD}$$
where the bottom morphism is a subobject inclusion and the left morphism is a differential of $F^p(C)$. Concretely, $A^{pq}_r$ is a pullback of $d^{p,p+q}$ along the subobject inclusion $F^{p + r}(C^{p + q + 1}) \subseteq F^p(C^{p + q + 1})$. Then $\ddot{A^{pq}_r}$ is the image of the composition in the following diagram
$$\begin{CD} A^{p - r + 1, q + r - 2}_{r - 1} @>>> F^{p - r + 1}(C^{p + q - 1}) \\ @VVV @VVV \\ F^p(C^{p + q}) @>>> F^{p - r + 1}(C^{p + q}) \end{CD}$$
and $\ddot{A^{pq}_{r + 1}}$ is the image of the composition in the following diagram
$$\begin{CD} A^{p - r, q + r - 1}_r @>>> F^{p - r}(C^{p + q - 1}) \\ @VVV @VVV \\ F^p(C^{p + q}) @>>> F^{p - r}(C^{p + q}) \end{CD}$$
To have a map from from one image to another, we need a map between their domains and codomains. But the pullback universal property only gives a morphism from $A^{p - r + 1, q + r - 2}$ to $A^{p - r, q + r - 1}$. Similarly, for the following screenshot
I only see how to construct a map from $A^{p + r, q - r + 1} \to A^{pq}_r$, for similar reasons.
So, my question is: is there is a mistake? If yes, can the proof be salvaged? If not, what am I missing?
We have $$\ddot{A^{p,q}_r}=\partial \newcommand\of[1]{\left({#1}\right)} \of{ F^{p-r+1}C^{p+q-1}\cap \partial^{-1} \of{ F^{p+1}C^{p+q} } }. $$ Note that $\partial^{-1}(F^{p+1}C^{p+q})$ is independent of $r$, and as $r$ increases, the filtration index decreases, and therefore the groups get larger, so you appear to be correct that the order of inclusions should be $\ddot{A}^{p,q}_r\subseteq \ddot{A}^{p,q}_{r+1}$.
This doesn't really affect the proof in any way (at least the visible part), because the excerpted discussion doesn't use the ordering at all. However, I will say that this ordering should be the correct one, since we want $Z^{p,q}_{r+1}\subseteq Z^{p,q}_r$ and $B^{p,q}_{r+1}\supseteq B^{p,q}_r$ so that $E_{r+1}^{p,q}$ is a subquotient of $E_r^{p,q}$.
$r$-cocycles should get smaller and $r$-coboundaries should get larger so that we have a sensible notion of convergence.
I imagine this was a typo, it's really easy to get indices and ordering confused here.