"Distribution" of numbers $0\leq a\leq b\leq c\leq d\leq 1$
Without the sorting, if you just choose $n$ random numbers $x_i$ such that $0 \leq x_i \leq 1$ and $\sum x_i = 1,$ the sample space can be represented as an $(n-1)$-dimensional polytope (specifically, a simplex) with vertices at $(1,0,0,\ldots,0)$, $(0,1,0,\ldots,0),$ $(0,0,1,0,\ldots,0), \ldots,$ $(0,0,\ldots,0,1).$ (That is, at each vertex exactly one coordinate is $1$ and the others are $0$.) An obvious choice of distribution over this region would be a uniform density.
I believe your unsorted differences achieve this distribution. Certainly they give the correct expected value for each variable.
Now you apply the restriction that $x_i \leq x_j$ if $i < j.$ The sample space for these variables is a smaller $(n-1)$-dimensional simplex that you can obtain by "cutting away" the parts of the original simplex in which $x_1 > x_2,$ or $x_2 > x_3,$ and so forth until you cut away the part in which $x_n > x_{n-1}.$
Generating the unsorted differences and then sorting them converts a uniform distribution over the larger simplex to a uniform distribution over the smaller one.
As an example, for $n = 3$ the final sample space has vertices at $(0,0,1),$ $\left(0,\frac12,\frac12\right),$ and $\left(\frac13,\frac13,\frac13\right).$ The centroid of this simplex is $\left(\frac19,\frac{5}{18},\frac{11}{18}\right),$ with coordinates in the ratio $2:5:11.$
For $n=4$ the vertices are $(0,0,0,1),$ $\left(0,0,\frac12,\frac12\right),$ $\left(0,\frac13,\frac13,\frac13\right),$ and $\left(\frac14,\frac14,\frac14,\frac14\right).$ For $n=5$ the vertices are $(0,0,0,0,1),$ $\left(0,0,0,\frac12,\frac12\right),$ $\left(0,0,\frac13,\frac13,\frac13\right),$ $\left(0,\frac14,\frac14,\frac14,\frac14\right),$ and $\left(\frac15,\frac15,\frac15,\frac15,\frac15\right),$ and so forth for larger $n.$
In short, each vertex is obtained by maximizing the value of $x_i$ for a different $i.$ It should be clear that for $n$ variables, the $i$th coordinate of the centroid of the final simplex, $i = 1, 2, \ldots, n,$ is $$ \frac1n\left(\sum_{j=1}^{n} \frac1j - \sum_{j=1}^{n-i} \frac1j\right). $$
Since summing up terms of the harmonic series can be a bit painful, here's a simpler way to generate the ratios. Denote the difference between the expected values of $x_{i+1}$ and $x_i$ by $\Delta_i = E(x_{i+1}) - E(x_i).$ Notice that if we take the expected value of $x_1,$ and follow it by the increasing sequence of these differences, we get a sequence of numbers in the ratio $$ E(x_i) : \Delta_1 : \Delta_2 : \cdots : \Delta_{n-3} : \Delta_{n-2} : \Delta_{n-1} = \frac1n : \frac{1}{n-1} :\frac{1}{n-2} : \cdots: \frac13 : \frac12 : 1. \tag{*} $$
In order to make the ratio on the right-hand side a ratio of integers, we merely need multiply each part by the least common multiple of $\{1,2,3,\ldots, n\}.$
For example, for $n = 5,$ we find that the least common multiple of $\{1,2,3,4,5\}$ is $60.$ The ratio $(^*)$ is therefore $$ \frac15 : \frac14 :\frac13 :\frac12 : 1 = \frac{60}{5} : \frac{60}{4} :\frac{60}{3} :\frac{60}{2} : 60 = 12 : 15 : 20 : 30 : 60. $$
Now, recalling that except for the first part of this ratio, each part is proportional to the difference between two consecutive expected values, and observing that $E(x_k) = E(x_1) + \sum_{i=1}^{k-1} \Delta_i,$ we replace the ratio with a ratio of partial sums: $$ 12 : 15 : 20 : 30 : 60 \to 12 : 27 : 47 : 77 : 137, $$ and there's the ratio you found.