Step-by-Step calculation of stationary wavelet transform gives shifted results?
Why normalized filters

Last question first, "Why $1/\sqrt2$": Because it makes the (Euclidean) norm of the filter $1$, so that the whole wavelet filter bank operation (if done right, that is, the decimated version) is orthogonal/isometric. It is a design choice, there is nothing wrong in staying with the integer values and correct the combined factor during reconstruction.

The book example is not really the standard wavelet transform

However, the book you cite is doing the wavelet transform undecimated. Which is a valid mathematical transform, but leaving out the decimation step, the down- and up-sampling, will lead to data bloat. By applying the two filters, the amount of data representing the signal is doubled, so some of that is redundant. This redundancy can be reduced by transmitting (or keeping for further processing) only every second element of these sequences.

Undecimated or otherwise redundant transforms give freedom in how to represent the signal, that is, in some non-linear post-processing one can set some of the data to zero and compensate (more-or-less) lossless-ly by changing other coefficients. However here I think this example is just a teaching aid.

Decimated version of the examples

So taking of cD1 and cA1 only every second entry gives cD1(2:2:end)=[0 0 0 0] and cA1(2:2:end)=[160 160 0 0], together they encode the same signal, which now has more zeros and is thus better suited for compression. In the reconstruction, zeros will be added to fill the gaps, and then the convolution with the reversed/biorthogonal filters is applied. So 

[1 -1]*[0 0 0 0 0 0 0 0] = [0 0 0 0 0 0 0 0 0]


and 

[1 1]*[160 0 160 0 0 0 0 0] = [160 160 160 160 0 0 0 0 0]


so that in the sum the result is twice the original signal, $(\sqrt2)^2$, with a zero added at the end.



If one shifts the whole process by one place, the reduced signals are [ -80 0 80 0] and [80 160 80 0] so that in the reconstruction one has to add

[1 -1]*[-80 0 0 0 80 0 0 0] = [-80 80 0 0 80 -80 0 0 0]


and

[1 1]*[80 0 160 0 80 0 0 0] = [80 80 160 160 80 80 0 0 0] 


to give [0 160 160 160 160 0 0 0 0]. Here we get, additional to the factor 2, also the shift, or an additional zero at the start of the sequence. However, by using the first, third etc. elements in down-sampling, we used implicitly an extension of the sequence S to the left by zero. This additional zero element is what is obtained now explicitly in the reconstruction.



In general wavelet transforms, the wavelet filters have lengths larger than 2, leading to more "interesting" boundary effects that include a shift by about the length of the filter.