Simple examples for the use of spectral sequences
The best example I can think of is the Lyndon-Hochschild-Serre spectral sequence in group cohomology. See for instance Chapter VII Section 6 of Brown's Cohomology of Groups.
The spectral sequence, for a group extension $1\to H\to G\to Q \to 1$ (I'm using Brown's notation) and for a G-module M, is of the form
$$E^2_{pq}=H_p(Q,H_q(H,M))\Rightarrow H_{p+q}(G,M).$$
Let's use this to calculate the third integral homology of the dihedral group D2n, for $n$ an odd integer. The group extension is $$1\to C_n\to D_{2n}\to C_2 \to 1,$$
and the corresponding Lyndon-Hochschild-Serre spectral sequence is
$$E^2_{pq}=H_p(C_2,H_q(C_n,\mathbb{Z}))\Rightarrow H_{3}(D_{2n},\mathbb{Z}),$$
where p and q add up to 3. The integral homology of a cyclic group Cm is $\mathbb{Z}$ for $q=0$, vanishes for $q\in\mathbb{N}^\ast$ even, and is $\mathbb{Z}/m\mathbb{Z}$ for $q$ odd. Plug this information into the Lyndon-Hochschild-Serre spectral sequence, and you find
$$H_{3}(D_{2n},\mathbb{Z})=H_0(C_2,\mathbb{Z}/n\mathbb{Z})\oplus H_3(C_2,\mathbb{Z})\simeq \mathbb{Z}/2n\mathbb{Z}.$$
This is easy and elegant calculation in my opinion, and occurs in practice in knot theory ($H_{3}(D_{2n},\mathbb{Z})$ turns out to be isomorphic to the relative bordism group of Fox n-coloured knots).
This is not the most profound answer but it's something that came up last week when I was reading a paper. The author wanted to prove that a certain obstruction to a problem was zero, and the obstruction lived in an $H^2$ that looked a bit scary: it was $H^2(W,V)$ with $W$ a local Weil group and $V$ a finite-dimensional vector space over a field of characteristic zero (I'll tell you all you need to know about this Weil group in a sec, in case you don't know what one is; it's a topological group coming up in number theory). But then I realised the obstruction was really in the image of $H^2(W/C,V)$ with $C$ a compact open subgroup of $W$ (a finite index subgroup of inertia).
But now I'm done because $W/C$ has a two-step filtration with a finite sub and a quotient isomorphic to $\mathbf{Z}$, and I know finite groups have no cohomology in char 0 in degrees 1 or more, and $\mathbf{Z}$ is the fundamental group of a 1-dimensional thing so it has no cohomology in degree 2 or more, and so by Hochschild-Serre, a calculation I can even do in my head in this example, there are no non-zero terms to build $H^2(W/C,V)$ from in $E_2$ and hence in $E_\infty$ so this group vanishes. I can do this calculation without even pulling out a piece of paper.
I'm sure if I were more "group-theoretic" I would have a much clearer picture about what was going on, but Hochschild-Serre just explains to me in a very concrete way how the cohomology of a group is built from the cohomology of its subs and quotients, and is definitely something I carry around in my "useful tools" bag.
This isn't exactly what you asked, but its a very simple example that (to me) demonstrates some of the necessity of the complexities of spectral sequences. Consider the ring $R=\mathbb{C}[x,y]$, and consider the module $M=(Rx+Ry)\oplus R/x$. Then the double dual spectral sequence converges to the original module: $$ Ext^{-i}_R(Ext^j_R(M,R),R) \Rightarrow M $$ The second page of this spectral sequence has
- $R$ in degree $(0,0)$
- $R/x$ in degree $(-1,1)$
- $R/(Rx+Ry)$ in degree $(-1,2)$
- A non-trivial knights-move map (differential on the second page) from $(0,0)$ to $(-1,2)$ which is the natural quotient map.
The spectral sequence collapses on the third page, with $(Rx+Ry)$ in degree $(0,0)$ and $R/x$ in degree $(-1,1)$. One shortcoming of this example is that you get the same module back, split apart into different components; rather than the associated graded of some interesting filtration. If memory serves, there was a way to tinker with this example to give it that property too, but it escapes me at the moment.