As far as I understood, there are two way to assign weights to a mixed Hodge complex (a way that does not depend on shifts, and another one that does). There is a clever way to relate these to ways using Deligne's decalage. Does there exist a conceptual explanation of these matters (including decalage)?
Unfortunately, I don't have a good conceptual explanation for any of this. So here's a somewhat more technical explanation of how decalage is used in Hodge theory. The actual business of constructing mixed Hodge structures tends to be quite involved. In Deligne's approach, the Hodge and weight filtrations $F$, $W$ are constructed directly on certain complexes $K,\ldots$ subject to a bunch of axioms that can be verified in natural situations. This datum is a so called mixed Hodge complex. One seemingly technical, but very important, consequence of these conditions is that:
The $E_1$ degeneration is a very natural condition, it means that $F$ is strictly compatible with differentials. It can viewed as an abstract generalization of the Hodge decomposition. (Note that this doesn't do away with harmonic theory, since one needs it to verify the above axioms in the first place.) However, the $E_2$ degeneration condition for $W$ is a bit harder to work with. Here decalage becomes very convenient, in that it converts $E_2$ degeneration to $E_1$ degeneration for a new filtration $Dec(W)$. This makes certain additional arguments much more manageable. In fact, this trick of passing to $Dec(W)$ is needed in the proof of the above theorem. Added Perhaps another instructive illustration of these ideas can be found in
Beilinson's paper Notes on absolute Hodge cohomology. In the paper, he introduces
a variant of a mixed Hodge complex called a $\tilde p$Hodge complex. Localizing the
category of these with respect to quasiisomorphism results in a triangulated
category $D^b_{\tilde H^p}$. He proves (or more accurately sketches a proof) that this is equivalent to the bounded derived category of polarizable mixed Hodge structures $D^b(H^p)$.
The functor This has got to be one least understandable answers that I've ever written here, but when you have complexes with several filtrations the story is going to be technical. The miracle is that it isn't worse than it is. 

