I am trying to understand the nuts and bolts of the spectral sequence associated to a double complex. Similarly to the question here: https://mathoverflow.net/questions/308514/construction-of-differentials-in-the-spectral-sequence-for-double-complexes, I understand that one can use the machinery of a filtered chain complex to construct this spectral sequence. However, I thought it would be instructive to also understand things at the "diagram chase" level.
The answer to the post above gives a direct construction of the $d^2$ differential with a diagram chase similar to the one used to prove the Snake Lemma. (This similarity is also pointed out in Ravi Vakil's notes https://math.stanford.edu/~vakil/0708-216/216ss.pdf).
Question: Can the $d^3$ differentials (and higher) be constructed similarly?
Thoughts: It seems to me a natural thing to try is the following. Let's say we have a double complex $C^{pq}$ with boundary maps of bidegree $(-1,0)$ and $(0,-1)$. I'll have the vertical maps as my $d^0$ differentials, so $d^0_{p,q} : E^0_{p,q} \to E^0_{p, q-1}$. Then the (horizontal) $d^1$ differentials are induced by the double complex structure maps, so $d^1_{p,q}: E^1_{p,q} \to E^1_{p-1,q}$ and the $d^2$ differential $d^2_{p,q}:E^2_{p,q} \to E^2_{p-2,q+1}$ can be constructed as in the post above.
Now let's say (for example and concreteness) we want to construct a map $d^3_{4,2}:E^3_{4,2} \to E^3_{1,4}$. Given $x \in E^3_{4,2}$, we can lift $x$ to an element $x' \in \ker(d^2_{4,2})$, so that $d^2_{4,2}(x') = 0$ in $E^2_{2,3}$. Therefore $d^2_{4,2}(x') \in \text{im}(d^1_{3,3})$, so let's say $d^1_{3,3}(y) = d^2_{4,2}(x')$.
Now if $y$ gave an element of $E^2_{3,3}$, we could consider $d^2_{3,3}(y) \in E^2_{1, 4}$ and define $d^3_{4,2}(x) = d^2_{3,3}(y)$, but I don't think $y$ is in the kernel of the relevant $d^1$ differential, so it doesn't actually give an element of $E^2$.
Is there some diagram chase that I'm not seeing? Or can the higher differentials not be constructed in such an elementary/obvious way? Any guidance would be appreciated!