The "correct" standard deviation

Before I answer your question: In general it is not true that the mean of two means is the mean of all points. consider the example $avg(avg(0,0,0),avg(1,1)) = 0.5 \neq 0.4 = avg(0,0,0,1,1)$.

Regarding the standard deviation: Only your first method actually makes sense, because the other methods do in general not coincide with the definition of the standard deviation.

If you consider your shrinkage estimates as samples from distributions with a common variance then the pooled estimate of the common variance is $$ s^2=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2} $$

In this expression you have a sample of size $n_1$ with sample variance $s_1^2$ and a sample of size $n_2$ with sample variance $s_2^2$

If I understand your data, you have $n_1=2$ for widget 1 and $n_2=2$ for widget 2 giving $$ s^2=\frac{s_1^2+s_2^2}{2} $$ so actually the variance is the average of the individual variances, in this case. The standard deviation is the square root of the variance.

This link may be helpful.