Strictly speaking, there is just one approach to a uniform proof, which is the one given by Elementary Differential Geometry , Christian Bär, pages 201-209. This approach is based on Riemannian geometry.
The impossibility of coming up with a 'rule-and-compass' uniform proof is that the Pythagorean theorem is expressed in essential different ways:
Euclidean geometry: $a^2+b^2=c^2$
Spherical geometry: $\cos(a)\cos(b)=\cos(c)$
Hyperbolic geometry: $\cosh(a)\cosh(b)=\cosh(c)$
It is true that we may derive the following formulae for rectangular triangles:
Euclidean geometry: $\sin(\alpha)=\frac{a}{c}$
Spherical geometry: $\sin(\alpha)=\frac{\sin(a)}{\sin(c)}$
Hyperbolic geometry: $\sin(\alpha)=\frac{\sinh(a)}{\sinh(c)}$
and then the usual proof of the sine rule applies to the other two cases (just dividing a triangle into two rectagle ones by an altitude), but whereas $\sin(\alpha)=\frac{a}{c}$ is a definition, the other two expressions have to be found in a different way.
For any point $P$ on unit sphere, let $P'$ be its antipodal point and $\hat{P}$ the corresponding unit vector. We will use the notation
$\mathcal{C}_{P_1P_2\cdots P_n}$ to denote a spherical arc starting from $P_1$, passing through $P_2,\ldots$ and end at $P_n$.
Let $A, B, C$ be any three points on unit sphere, close enough to fit
within half of a hemisphere
(i.e. a spherical lune of angle $\frac{\pi}{2}$ ).
Let $\Omega_{ABC}$ be the area of spherical $\triangle ABC$.
It can be computed using a formula by Oosterom and Strackee
$$\tan\left(\frac{\Omega_{ABC}}{2}\right) = \frac{ \left|\hat{A}\cdot (\hat{B} \times \hat{C})\right|}{1 + \hat{A}\cdot\hat{B} + \hat{B}\cdot\hat{C} + \hat{C}\cdot\hat{A}}$$
In the special case where $A, B$ lies on the equator, symmetric with respect to $x$-axis and $C$ lies on the upper hemisphere, i.e.
$$
\begin{cases}
A &= (\cos\alpha,-\sin\alpha,0),\\
B &= (\cos\alpha,+\sin\alpha,0),\\
C &= (x, y, z)\end{cases}
\quad\text{ where } \alpha \in (0,\frac{\pi}{2}), z > 0$$
Above formula reduces to
$$\tan\left(\frac{\Omega_{ABC}}{2}\right) = \frac{\sin\alpha z}{\cos\alpha + x}$$
This implies the locus of $P$ in upper hemisphere for fixed
$\Omega_{ABP} = \Omega_{ABC}$ is the circular arc $\mathcal{C}_{B'CA'}$.
Let $\theta$ be the angle between $\mathcal{C}_{B'CA'}$ and $\mathcal{C}_{B'ABA'}$ at $B'$. The plane holding the locus has the form
$$t ( c + x ) - s z = 0\quad\text{ where }\quad
\begin{cases}
t &= \tan\left(\frac{\Omega_{ABC}}{2}\right)\\
c &= \cos\alpha\\
s &= \sin\alpha
\end{cases}$$
The normal vector of the plane is pointing along the direction $(t, 0, -s)$. This means the tangent vector of $\mathcal{C}_{B'CA'}$ at $B'$ is along the direction $(t, 0, -s ) \times ( -c, -s, 0 ) \propto (s, -c, t)$.
Notice the tangent vector of $\mathcal{C}_{B'ABA'}$ at $B'$
is pointing along the direction $(s,-c,0)$, we find
$$\cos\theta = \frac{s^2 + c^2 + 0}{\sqrt{s^2+c^2}\sqrt{s^2+c^2+t^2}} = \cos\left(\frac{\Omega_{ABC}}{2}\right)$$
From this, we can deduce the circular arcs
$\mathcal{C}_{B'PA'}$ and $\mathcal{C}_{B'ABA'}$ intersect at an angle $\frac{\Omega_{ABC}}{2}$.
This leads to following construction of the desired "spherical centroid" $X$.
Construct the circular arcs $\mathcal{C}_{B'CA'}$ and $\mathcal{C}_{B'ABA'}$,
Trisect the angle $\angle AB'C$ - i.e. find a circular arc
$\mathcal{B'\tilde{C}A'}$ such that $\tilde{C}$ is lying on same side as $C$ with respect to $AB$ and $\angle AB'\tilde{C} = \frac13 \angle AB'C$.
Repeat above procedures for other two sides of $\triangle ABC$ to get
circular arcs $\mathcal{C}_{C'\tilde{A}B'}$ and $\mathcal{C}_{A'\tilde{B}C'}$.
$X$ will be lying on the common intersection of the three circular arcs
$\mathcal{C}_{B'\tilde{C}A'}$, $\mathcal{C}_{C'\tilde{A}B'}$ and $\mathcal{C}_{A'\tilde{B}C'}$.
Best Answer
In the Law of Sines for spherical triangles, $$ \frac{\sin a}{\sin A} = \frac{\sin b}{\sin B} = \frac{\sin c}{\sin C} = \frac{\sin a \sin b \sin c} {\sqrt{1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c}}. $$ The formula on the far right bears some vague resemblance to a formula for the corresponding constant in plane geometry, but I do not have a "nice" interpretation of it.
This notation assumes that the "lengths" of the sides are measured by the angle each side subtends at the center of the sphere. (If the sphere has radius $1$, this is the same as the length of the arc forming each side.)
The relationship between the circumradius of the triangle and the ratio $\frac{\sin a}{\sin A}$ is not going to be as simple as the corresponding relationship for plane triangles, as a simple example will demonstrate. Consider the triangles $\triangle ABC$ and $\triangle AB'C'$ with the following spherical coordinates: $$ \begin{eqnarray} \phi &=& 0 &\mbox{ at } A \\ (\phi.\theta) &=& \left(\frac\pi4,0\right) &\mbox{ at } B \\ (\phi.\theta) &=& \left(\frac\pi4,\frac\pi2\right) &\mbox{ at } C \\ (\phi.\theta) &=& \left(\frac{3\pi}4,0\right) &\mbox{ at } B' \\ (\phi.\theta) &=& \left(\frac{3\pi}4,\frac\pi2\right) &\mbox{ at } C' \end{eqnarray} $$
That is, $A$ is on the axis from which $\phi$ is measured; $B$ and $B'$ are on a common great circle through $A$; and $C$ and $C'$ are on a common great circle through $A$. Clearly both triangles have the same angle at $A$, and by symmetry the length of the side opposite $A$ is the same in both triangles; that is, if the arc from $B$ to $C$ is $a$ and from $B'$ to $C'$ is $a'$, then $a = a'$ and $$\frac{\sin a}{\sin A} = \frac{\sin a'}{\sin A}.$$ But these triangles clearly do not have congruent circumcircles, so if $\frac{\sin a}{\sin A}$ equals some function of the circumradius of $\triangle ABC$, it follows that $\frac{\sin a'}{\sin A}$ does not equal the same function of the circumradius of $\triangle AB'C'$.
Some interesting facts about this are shown by Odani, Kenzi. "On a Relation between Sine Formula and Radii of Circumcircles for Spherical Triangles." 愛知教育大学研究報告. 自然科学編 59 (2010): 1-5. https://aue.repo.nii.ac.jp/?action=pages_view_main&active_action=repository_view_main_item_detail&item_id=633&item_no=1&page_id=13&block_id=21 One of these facts is that if $R$ is the circumradius of $\triangle ABC$, then $$ \left(\frac{\sin a}{\sin A}\right)^2 = 4 (\tan^2 R) \left(\cos^2\frac a2\right) \left(\cos^2\frac b2\right) \left(\cos^2\frac c2\right). $$ An immediate consequence of this is that for any non-degenerate spherical triangle $\triangle ABC$, $$ \frac{\sin a}{\sin A} < 2 \tan R. $$ But for "most" triangles, the ratio $\frac{\sin a}{\sin A}$ is not close to this limit. If $\triangle ABC$ is a large triangle whose circumcenter lies within the triangle, the upper bound of $\frac{\sin a}{\sin A}$ is quite a bit smaller than this. The reference above is quite explicit about what the upper (and lower) bounds of the ratio are; there are four separate cases to consider, and in each case the upper and lower bounds are functions of the circumradius, but not always as simple as $2\tan R$.