I like the account in Len Evens' book `The cohomology of groups'. It sets up the Lyndon-Hochschild-Serre spectral sequence using the tensor product of a free $\mathbb{Z}Q$-resolution for $\mathbb{Z}$ and a free $\mathbb{Z}G$-resolution for $\mathbb{Z}$: similar to the proof that Fernando mentions but using arbitrary resolutions. I find that double complexes are easier to handle than the more general filtered complexes.

Sorry I didn't reply immediately; I wanted to think up a good answer. This stuff is covered in Wolfgang Luck's Transformation Groups and Algebraic K-Theory, but I think I learned it by looking at the proof of the Eilenberg-Ganea theorem in Ken S Brown's book `Cohomology of Groups'. I'm not sure that these references will be what you want, but I can't think of a better one.

Yes, you've got it.

Because $\pi_2(\widetilde X)$ is a free module for $\pi_1(K)\cong\pi_1(X)$, you choose the correct quantity of 2-cells to attach to $K$ to produce $L$. E.g., if $\pi_2(X)$ is free of rank 2, then you attach two 2-cells to $K$. In this way you ensure that $\pi_2(L)\cong \pi_2(X)$ as modules for $\pi_1(L)=\pi_1(K)\cong \pi_1(X)$. At this point you still have not defined the map $\psi$ on the 2-cells of $L$; you choose $\psi$ to realize some isomorphism between $\pi_2(L)$ and $\pi_2(X)$. By construction $\tilde\psi: \pi_2(\widetilde L)\rightarrow \pi_2(\widetilde X)$ is isomorphic.

Now that I've understood the problem, I think I can answer your question: see below.

Silly me: you are right. Of course the 3-sphere does not satisfy condition $\mathcal{D}_2$. The idea is that a complex that satisfies $\mathcal{D}_2$ should be quite close to being 2-dimensional.

That is exactly my point: if $X=S^3$, then for any 2-complex $L$, there is only one homotopy class of maps from $L$ to $X$, and so the mapping cylinder of any $\psi:L\rightarrow X$ must be homotopy equivalent to $X=S^3$, and so no such $\psi$ can be 3-connected. There must be an extra hypothesis that $X$ is 2-dimensional, otherwise the result is not true.

Surely there must also be a hypothesis like `$X$ is 2-dimensional'? The 3-sphere $S^3$ (whose fundamental group is free of rank zero) seems to give a counterexample otherwise.

I have put up what I wrote. I'm sure that most books about $G$-CW-complexes will contain the result, but I don't know of a good reference offhand.

If you do not require the complex to be spherical, then any $G$-representation can be realized as $H_1$ of a 2-complex, or as $H_2$ of a simply-connected 3-complex - I got as far as writing this out in detail as an answer before noticing the condition `spherical'.