For the completeness of the discussion, i must add a reconstruction of Gauss's procedure, which i discovered a few months ago. I already posted it on HSM stack exchange, so i simply copy it to here:
Gauss's procedure does imply Bolyai's result on the volume of orthoscheme tetrahedron, as i'll show here. However, Gauss's result is a little bit more limited than Bolyai, since in Gauss's tetrahedron 4 of the 12 face angles of the tetrahedron are right, while Bolyai refers to a slightly more general tetrahedron whose only 3 face angles ar right.
Preliminary discussion: To see the connection between Schlafli formula and the first formula in Gauss's fragment, one needs to understand that Gauss thinks of the tetrahedron 1234 in such a way that the faces 124 and 134 are perpendicular and the edges 24 and 13 meet the intersection line 14 also at right angles. In addition, Gauss defines the tetrahedron in such a way that the angles at vertex 3 are constant (so that an "observer" in hyperbolic space which is located at vertex 3 sees the rest of the vertexes at constant lines of sight).
Therefore, only the dihedral angle of the side 24 changes. This leads directly to the first formula in Gauss's fragment (apart from a missing factor of $\frac {1}{2}$).
Derivation of explicit volume formula from Gauss's formula:
For the sake of consistency, we denote the angles 431, 234, and 214 as $\alpha$, $\beta$ and $\gamma$, respectively. Now lets look at the link of vertex 3 of the tetrahedron: it's a spherical triangle whose two edges lengths are $\alpha$ and $\beta$ and one angle is $\gamma$ (it is the dihedral angle of edge 31 and it's also equal to $\gamma$). In addition the sides $\alpha$$, \beta$ of this spherical triangle are orthogonal to each other. Therefore, by a combination of the spherical sine theorem and the spherical pythagoras theorem, we get:
$$ \frac{{sin(arccos(cos\alpha\cdot cos\beta))}}{{sin 90}} = \frac {{sin\beta}}{{sin\gamma}}$$, or:
$$sin\gamma = \frac {{sin\beta}}{{\sqrt{{1 - (cos\alpha \cdot cos\beta)^2}}}}$$
Now, Gauss's procedure for the calculation of the volume leads to the following integral:
$$\Delta = \frac {{tan\beta}}{{2 tan \gamma}}\int_{0}^{c}\frac {{x sinh(x) dx}}{{(cosh^2(x) - 1 + \frac {{cosh^2(x)}}{{sin^2\alpha cot^2\beta}})\sqrt{{\frac{{cosh^2x}}{{cos^2\beta}} - 1}} }}$$.
Now, the left factor of the denominator $cosh^2(x)(1 + \frac{{1}}{{sin^2\alpha \ cot^2\beta}})-1$, is exactly equal to $cosh^2(x)\cdot \frac{{1}}{{cos^2\gamma}}-1$, because subtitution of $cos\gamma = \sqrt {1 - \frac {{sin^2\beta}}{{1-(cos\alpha\cdot cos\beta)^2}}}$ in this expression gives the previous one.
Concluding remarks:
- As can be seen from this presentation - Hyperbolic Volumes and Symmetry, the Bolyai's volume integral (see Theorem 5, p. 12, at this presentation):
$$Vol(T) = \frac {{tan\beta}}{{2 tan \gamma}}\int_{0}^{c}\frac {{x sinh(x) dx}}{{(\frac {{cosh^2(x)}}{{cos^2\gamma}} - 1)\sqrt{{\frac{{cosh^2x}}{{cos^2\beta}} - 1}} }}$$
can be seen as treating a slightly more general case then the one treated by Gauss (note: the differences in notation between the Bolyai integral in the presentation and Gauss's integral are just due the different symbols of the angles 431, 234, and 214).
However, for the case treated by Gauss, his formulas are absolutely correct. He should also be given credit for the identification of the calculation of the orthoscheme tetrahedron as the basis for volume formulas of general tetrahedrons (without right angles). In one of his letters, he refered to those calculations of volumes as "die jungle" - i guess he refered to the extremely complicated integrals that arise in the attempts to the decompose the general tetrahedron into orthoscemes (this problem was only solved very recently).
- Paul Stackel, the mathematician who edited Janos Bolyai's geometric works, had the following things to say about Bolyai's derivation of his integral formula:
It is most remarkable that the method that Gauss used for cubing the tetrahedron, is exactly the same as that of Johann. This is shown in a note from March 1832, from Gauss's estate, which is printed in the works (vol. VIII, p. 228); Gauss has exactly the same special tetrahedron (only 3142 instead $abc\delta$ means) and exactly the same decomposition by planes perpendicular to ab (31).
This quotation is taken from p. 113 of the book "Wolfgang und Johann Bolyai geometrische Untersuchungen" (here is a link: https://archive.org/details/wolfgangundjohan01stuoft/page/112/mode/2up), which was edited and translated to german by the Stackel.
Since it was the same Stackel who also pointed out that Bolyai discovered a form of Schlafli's formula, i think it confirms the view that Gauss really discovered a special case of Schlafli formula. This is a very significant point in Gauss's work on non-euclidean geometry that went unnoticed and was never commented on seriously by historians of mathematics. I think so because it's perhaps Gauss's only fragment dealing with calculations in hyperbolic space $H^3$, and not just in the hyperbolic plane.
- It's still necessary to understand how Gauss arrived at the formula $\partial \Delta = -\frac{{1}}{{2}}(24)d(341)$ (he missed the factor $\frac {{1}}{{2}}$ at the first attempt); the second formula from his note can be derived with relative ease. I think dechipering Gauss's second fragment might serve as a clue for understanding Gauss's reasoning.