In Weibel (Introduction to Homological Algebra)'s proof that left derived functors form a homological $\delta$-functor (Thm. 4.2.6), he does a lot of work that seems unnecessary to me. The relevant pages can be seen on Google Books:

Once he's established the SES $0\to F(P')\to F(P)\to F(P'')\to0$, aren't the $\partial$s automatically natural (Prop. 1.3.4)?

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Natural in what sense? They depend on the short exact sequence of resolutions a priori. You now need to refine the comparison theorem of resolutions and the horseshoe lemma to show that you can build a map of these short exact sequences of resolutions and that this map is unique, and that's precisely what Weibel does. – t.b. Jul 22 '11 at 12:01

You are probably confused because Proposition 1.3.4 proves that given a short exact sequence $0\to A\to B\to C\to 0$ of chain complexes there is an associated long exact sequence in homology, and for each map of short exact sequences, there is an associated map of long exact sequences, and in particular the connecting homomorphism is natural in the sense that it commutes with the map of long exact sequences.
The thing is, we don't have a map between the SESs of chain complexes that were used to give the long exact sequence for the derived functors. All we have is the map between the SESs of objects in the Abelian category and we use it to produce a map between the chain complexes $0\to F(P')\to F(P)\to F(P'')\to 0$ and $0\to F(Q')\to F(Q)\to F(Q'')\to 0$, and then that gives the naturality of the connecting map.
A possible source of confusion is the following: While it is true that we can choose $P = P' \oplus P''$ by the Horseshoe lemma, it is not true that this gives that the resolution $P \to A$ obtained from it is the direct sum of the resolutions $P' \to A'$ and $P'' \to A''$ unless $A = A' \oplus A''$ (the differentials are different). – t.b. Jul 23 '11 at 8:53
Let me see if I understand you correctly: while we can obtain a chain map $F\colon P\to Q$ by the Horseshoe lemma, we don't automatically have $Fd=dF'$ and $F''d = dF$; and thus we need to carefully construct $F$ explicitly? – Yuri Sulyma Jul 26 '11 at 19:26