Is there a book (A User's Guide to Spectral Sequences? Brown's Cohomology of Groups?) that goes through a calculation of the (co)homology of the Klein bottle using a spectral sequence, treating it as a fiber bundle with base $S^1$ and also fiber $S^1$?
What I really need to compute is the (co)homology of a fiber bundle with base $T^2$ (the standard 2-torus) and fiber the Brieskorn homology sphere $\Sigma(3,4,5)$ using trivial outer action but a non-trvial element of $H^2(Q; Z(K))$ in the total group of the fiber bundle $E$'s fundamental group, but I'd like to start with a problem where I know the answer, as I've never done a spectral sequence calculation before.
I am wholly ignorant of the literature, though I have heard Hatcher's notes are good to play with for practice. I will simply run through the computation for you. I apologize if I draw the SS differently than others, I forget sometimes how most people orient it --- you may have to swap the axes.
What you know is that there is a spectral sequence $$H^*(B;H^*(F;\Bbb Z)) \implies H^*(E;\Bbb Z).$$ This should be interpreted carefully: this is cohomology with local coefficients, where $H^*(F;\Bbb Z)$ is a $\pi_1 B$-module.
Remember that the Klein bottle may be defined as $K = S^1 \times [0,1]/(z,0) \sim (\overline z, 1)$, gluing the boundaries together by a reflection of the circle. This is in fact how the fiber bundle structure is defined; the projection to the base is the projection to the second factor $K \to [0,1]/0 \sim 1$, which is homeomorphic to a circle. Thus the monodromy action of $\pi_1 B = \Bbb Z$ is such that $1$ acts as reflection on the circle. The induced map of reflection on cohomology is the identity on $H^0(S^1)$ and $-1$ on $H^1(S^1)$.
Thus our spectral sequence starts with bottom line $H^*(S^1;\Bbb Z)$ (which is a copy of $\Bbb Z$ in degrees zero and one) and top line $H^*(S^1;\Bbb Z_-)$ (which is a copy of $\Bbb Z/2$ in degree one and nothing else). I encourage you to do the local coefficient computation on your own.
So our spectral sequence starts from $E_2$ page $$0 \; \;\;\;\;\; 0 \; \; 0 \; \;\; \;$$ $$0 \; \; \Bbb Z/2 \; \; 0 \; \; \cdots $$ $$\Bbb Z \; \; \Bbb Z \; \; 0 \; \; \cdots$$
But all differentials from here on out have bidegree $(k, 1-k)$, and since we are at the $E_2$ page they will all have first degree $k \geq 2$ --- so that the differentials are all automatically zero. Thus the spectral sequence collapes at $E_2$.
We thus obtain $H^0(K; \Bbb Z) = \Bbb Z$, as well as $H^1(K;\Bbb Z) = \Bbb Z$ and $H^2(K;\Bbb Z) = \Bbb Z/2$. You may check this is the same thing you would get from, say, Mayer-Vietoris.