Here's a little demo to see what projective transformations of a triangle to itself look like (and what they do to its circumcircle). $\alpha$ is the scale factor on the rightmost vertex, $\beta$ is the scale factor on the top-left vertex and $\gamma$ is the scale factor on the bottom-right vertex.
Let $(T, \ell)$ be a triangle mesh with a discrete metric. We can equip this mesh with a canonical hyperbolic metric with cusps as follows:
Consider a euclidean triangle with its circumcircle. If we interpret the interior of the circumcircle as a hyperbolic plane in the Klein model, then the euclidean triangle becomes an ideal hyperbolic triangle, that is, a hyperbolic triangle with vertices at infinity. This construction equips any euclidean triangle (minus its vertices) with a hyperbolic metric. If it is performed on all triangles of a euclidean triangulation $(T, \ell)$ then the hyperbolic metrics induced on the individual triangles fit together so $T \setminus V$ is equipped with a hyperbolic metric with cusps at the vertices. Thus, $T$ becomes an ideal triangulation of a hyperbolic surface with cuspsBobenko et al.
To prove this proposition, we start with a useful lemma.
We will work in homogeneous coordinates. The circumcircle is a squished cone in $\R^3$, so we can represent it as the 0-set of a quadratic form. A circle in $\R^2$ is determined by the equation \[(x+c)^2 + (y+d)^2 - r^2 = 0\] Multiplying this out, we obtain \[x^2 + y^2 + 2cx + 2dy + c^2 + d^2 - r^2 = 0\] When we pass to $\R^3$, we want the polynomial to be homogeneous (since it must be invariant under scaling). So on $\R^3$, we have \[x^2 + y^2 + 2cxz + 2dyz + (c^2 + d^2 - r^2)z^2 = 0\] If we let $e := c^2 + d^2 - r^2$, we can just write this as \[x^2 + y^2 + 2cxz + 2dyz + ez^2 = 0\] This is linear in $c,d$ and $e$, so we can use the fact that our cone passes through the 3 points of our triangle to determine $c$, $d$, and $e$ uniquely.
Let $w_1, w_2, w_3$ be the (homogeneous) coordinates for the verticles of the original triangle and $\tilde w_1, \tilde w_2, \tilde w_3$ be the (homogeneous) coordinates of the updated triangle. Since we want a projective map identifying the first triangle with the second, we have to send $w_i \mapsto a_i \tilde w_i$, but we have the freedom to choose the scale factors $a_i$. Note that since our vertices are linearly independent, the choice of the $a_i$ determines our projective map.
We will show that there is a unique choice of the $a_i$ such that our map preserves the triangle's circumcircle. Let $q$ be the quadratic form representing the original triangle's circumcircle, and $\tilde q$ be the quadratic form representing the updated triangle's circumcircle. It will be convenient to consider the bilinear forms $b, \tilde b$ associated with the quadratic forms $q, \tilde q$. A map $f:\R^3 \to \R^3$ preserves circumcircles if $q(x)$ equals $\tilde q(f(x))$ up to a scale factor, i.e. \[q(x) = \mu \; \tilde q(f(x))\] In terms of the bilinear form, this means that \[b(x, y) = \mu\;\tilde b(f(x), f(y))\] Since the $w_i$ form a basis of $\R^3$, this is true if and only if it holds on the basis vectors \[b(w_i, w_j) = \mu\;\tilde b(a_i \tilde w_i, a_j\tilde w_j) = \mu a_i a_j \; \tilde b(\tilde w_i, \tilde w_j)\] Furthermore, note that the $z$-component of $w_i-w_j$ is 0. Thus, \[q(w_i-w_j) = \ell_{ij}^2\] Using the fact that $b(w_i, w_i) = 0$, we find that \[\begin{aligned} \ell_{ij}^2 &= q(w_i-w_j)\\ &= b(w_i-w_j, w_i-w_j)\\ &= -2b(w_i, w_j) \end{aligned}\] Thus, we conclude that our map preserves circumcircles if and only if \[\ell_{ij}^2 = \mu a_i a_j \tilde \ell_{ij}^2\] So the circumcircle-preserving projective map determined by the $a_i$ is equivalent to having conformal scale factors \[e^{u_i} = \mu^{-1/2}a_i^{-1}\]
Here is a demonstration of this proof.
Now we can prove the proposition
The proof is a quick computation which is probably easier to do than read.
This means that we can push all of the rescaling to the beginning (or end). Then we just have to do the flips in order.