@Alexis [feel free to ignore this bit - it's basically just me reminding myself of how the test works] If xᵢ, i=1,...,n are the data and yᵢ are the corresponding order statistics, and mᵢ are a set of corresponding S-W scores, the Shapiro-Wilk test consists of a numerator which is the square of m'y and whose denominator is the sums of squared deviations of the x's from their sample mean (n-1 times the usual variance of x and perforce, n-1 times the variance of y).
Algebraically, the ratio is always ≤1. When the order statistics are for a sample from the normal distribution, the numerator is generally very close in size to the denominator. It's akin to a squared correlation between order statistics and their expected value (except that the Shapiro Wilk scores incorporates the variance-covariance matrix of standard normal order statistics, in a manner akin to a standardized Mahalanobis distance); in small samples this will improve the power,
I think its impact reduces in very large samples though (it's a while since I looked at that stuff). . . . . . . . . [end of bit to ignore]
When you have a sample that is from a distribution which is different from the normal, the yᵢ values (the order statistics) are far from their expected values, reducing the correlation with the S-W scores and the typical size of the numerator decreases in relative to the variance on the denominator.
That is, you reject only for small values of W.
It doesn't matter what you do with W after that, up to monotonic transformation; all the values in the rejection region will be up one end of the distribution.
It converts deviations from normality, in all their multifarious dimensions, into a single direction of deviation from it, rather like an F statistic in one-way ANOVA converts all the various patterns of deviations in means from equality into a single statistic that will tend to be larger than it is under equality, whatever that pattern may be.
As a result, the test in either case (F in ANOVA, or the Shapiro Wilk) is inherently "one sided".
(Similar comments apply to the Shapiro-Francia test)
It doesn't really matter if you convert W into some Z-score, your rejection region (corresponding to values of W inconsistent with normality) will still only be in one tail.
Values in the other tail are the ones most consistent with normality, in the same way that a very small F-value in an ANOVA should not lead you to conclude that the means differ.
Analogously, I could readily convert (via monotonic transformation) F to a Z-score, but I should still only conclude the means differ for the tail of that Z which corresponded to a large F.