For this answer I will not rely on Cartesian coordinates, but I will not work directly in terms of latitude and longitude either. Instead, it will be convenient to represent a point with spherical coordinates $(\phi,\theta)$ as $z=e^{i\phi}\tan\dfrac\theta2$, corresponding to a stereographic projection of the sphere into the complex plane. We can now apply Mobius transformations to move the complex numbers $\alpha$, $\beta$ (corresponding to $A$, $B$) to more helpful locations.
Specifically, we can use rotations to move $\alpha$ to zero, then map $\beta$ to the real line, and finally rotate $\{\alpha,\beta\}$ to $\{r,r^{-1}\}$ on the positive real line. This amounts to rotating our coordinates so that $A$ and $B$ have zero longitude and opposite latitudes. To make this simpler, we shall assume that $\alpha$ is real without loss of generality (i.e. we take the longitude of $B$ relative to $A$).
We now state these transforms explicitly. First, note that the subgroup of the Mobius group corresponding to rotations consists of mappings of the form $z\mapsto \dfrac{u z-\overline{v}}{v z+\overline{u}}$ for complex $u,v$. A mapping of this form which takes $\alpha$ to the origin is $z\mapsto \dfrac{z-\alpha}{\alpha z+1}$; this also maps $\beta\mapsto \dfrac{\beta-\alpha}{1+\beta\alpha}= R e^{i\psi}$ for some real $R,\psi$. A rotation $z\mapsto e^{-i\phi} z$ then brings $\beta$ to $R$.
For the last transformation, consider the mapping $z\mapsto \dfrac{1+z r}{r-z}$ for real $r>0$ which takes the real line to itself. Then $\{0,R\}\mapsto \left\{\dfrac{1}{r},\dfrac{1+r R}{r-R}\right\}$; since we want the images of $0$ and $R$ to be on equal and opposite sides of the 'equator', we require their product to be unity. From this we deduce $r=R+\sqrt{1+R^2}$. Composing the three transformation gives the net rotation
$$\mathcal{T}:\;z
\mapsto \dfrac{z-\alpha}{1+\beta\alpha}
\mapsto e^{i\phi}\left(\frac{z-\alpha}{1+\beta\alpha}\right)
\mapsto \dfrac{1+e^{i\phi}\left(\frac{z-\alpha}{1+\beta\alpha}\right)r}{r-e^{i\phi}\left(\frac{z-\alpha}{1+\beta\alpha}\right)}
=\dfrac{e^{-i\phi}(1+\beta\alpha)+(z-\alpha)r}{r e^{-i\phi}(1+\beta\alpha)-(z-\alpha)}$$
We now can give a criterion for the original question: a point $z$ lies between $\alpha$ and $\beta$ if $r^{-1}\leq | \mathcal{T}(z)|r$. In principle one can do inverse transformations to map this annulus back to the original coordinates, but I won't do so explicitly. Instead I'll just summarize the result:
Let $\alpha,\beta$ be the stereographic coordinates of two points $A,B$ on the unit sphere, and define $w=R e^{i\phi}:=\dfrac{\beta-\alpha}{1+\beta\alpha}$. Then a point $C$ on the unit sphere lies between $A$ and $B$ iff its sterographic coordinate $z$ satisfies
$$\frac{1}{r}< \left|\dfrac{e^{-i\phi}(1+\beta\alpha)+(z-\alpha)r}{r e^{-i\phi}(1+\beta\alpha)-(z-\alpha)}\right|<r$$ where $r=R+\sqrt{1+R^2}$.
In the following, we consider that we are working on a unit sphere ($\|V_1\|=\|V_2\|=1$).
If your longitude and latitude coordinates are $(\theta_k,\lambda_k) \ k=1,2.$ Let
$$V_1=\begin{bmatrix}\cos \theta_1 \cos \lambda_1 \\ \sin \theta_1 \cos \lambda_1 \\ \sin \lambda_1 \end{bmatrix} \ \ \ V_2= \begin{bmatrix}\cos \theta_2 \cos \lambda_2 \\ \sin \theta_2 \cos \lambda_2 \\ \sin \lambda_2 \end{bmatrix}$$
Set $V_3=V_1+V_2$, and $V_4=\dfrac{V_3}{\|V_3\|}$: it is the desired point.
If one desires the spherical coordinates
of $V_4$ out of its cartesian coordinates $(a,b,c)$, it suffices to set :
$$\lambda = \text{asin}(c) \ \ \ \text{and} \ \ \ \theta = \text{asin}(\dfrac{b}{\cos \lambda})$$
Remark : if one wants a parametrized equation for the great circle passing through $V_1$ and $V_2$, here is a classical way to obtain it (cometimes called "orthonormalization"). It suffices to obtain an orthogonal basis of plane $(P)$ defined by $V_1$ and $V_2$. This is easily done by taking as first vector $V_1$ and as a second vector a linear combination $V=\alpha V_1+\beta V_2$ such that $\|V\|^2=1$ and $V \perp V_1 \iff V \cdot V_1=0$, generating two equations with two unknowns (let us recall that we have assumed that $\|V_1\|=\|V_2\|=1$):
$$\begin{cases}\alpha^2 +\beta^2+ 2 d \alpha \beta &=&1\\\alpha+d \beta=0\end{cases}\tag{1}$$
(where $d$ is defined as the dot product $d:=V_1\cdot V_2$) which is amenable to a quadratic equation with two solutions (in general) because its discriminant is $>0$.
Let $W_1=\alpha V_1+\beta V_2$ be one of these two solutions ; the looked for parametric equation of the great circle passing through $V_1$ and $V_2$ is
$$V = \cos(t) V_1 + \sin(t) W_1$$
Best Answer
To find the angular distance between two points, defined by two angles each, is a classical problem in spherical astronomy and in geography.
If the two coordinates are: $\phi=$right ascension (longitude in geography) and $\theta=$ declination (latitude). Than the angular distance $\alpha$ between two points $A,b$ is given by ( here a proof Great arc distance between two points on a unit sphere, but note that here the angle $\theta$ is that used in spherical coordinates, so it is the complement of the latitude): $$ \cos \alpha= \sin \theta_A\sin \theta_B+\cos \theta_A\cos \theta_B\cos(\phi_A-\phi_B) $$
We can use this formula to solve the problem in OP.
Let $(\phi_C,\theta_C)$ the coordinate of the center of a circle on a spherical surface. The points $P$ of the circle have coordinates $(\phi,\theta)$ such that the angular distance between $P$ and $C$ is a constant value $\rho$ (it is the quotient between the radius of the circle and the radius of the sphere), so the equation of the circe can be written as: $$ \cos \rho=\sin \theta\sin \theta_C+\cos \theta\cos \theta_C\cos(\phi-\phi_C) $$
Now we note that this equation contains three parameters $\rho,\phi_C,\theta_C$ so, in principle, we can find this parameters substituting the coordinates of three points for $\phi$ and $\theta$. I never performed these calculations, but they seems a bit complexes. I suppose that the simpler way to solve the problem in OP is to change to cartesian coordinates and to find the intersection between the plane determined by the three points and the sphere, as suggested in the comments.