Why does Klein's quartic curve have 24 heptagonal faces/56 triangular faces as a tiling? I first learned about Klein's quartic from John Baez' blog:
https://math.ucr.edu/home/baez/klein.html
In it, he claims that if we want to wrap up the heptagonal tiling of the hyperbolic plane by
"we can only do this if we use precisely 24 heptagons"... but why?  Why couldn't I take say, 48 heptagons instead?  Or 86? Or any other number so long as both sides of Euler's formula works out to integer values?
Everywhere I've looked for reading material on Klein's quartic as a "compactified tiling" of the hyperbolic plane, (including the book The Eightfold Way) seems to assume how many faces we have to compute the genus, or assumes the genus to compute the number of faces...
I believe that if one makes the connection of this tiling to the modular curve $\mathbb{H}/\Gamma(7)$, you can simply do some group theory/geometry to find 24 cusps, and make some argument from there about the genus etc.  However, I'm interested in carrying out this procedure for more general hyperbolic tilings of the plane by triangles.  Is there some way to see 24 heptagons as the "natural" way to compactify this tiling?
Suppose I tile the hyperbolic plane by triangles with angles $\frac{\pi}{2}, \frac{\pi}{m}, \frac{\pi}{n}$.  (where I'm assuming that $m$ and $n$ are sufficiently large the tiling is indeed hyperbolic, and not spherical or Euclidean) Now I want to "curl this up" like one curls up the tiling  of the hyperbolic plane by $\frac{\pi}{2}, \frac{\pi}{3}, \frac{\pi}{7}$ to get Klein.  How do I figure out what the natural choice is for the number of faces?
Once I've done that, how do I determine the gluing instructions?
 A: Recall Euler's formula: given a $2$-manifold $M$ tiled using $V$ vertices, $E$ edges and $F$ faces, its Euler characteristic is $\chi(M) = V - E + F$.
Suppose your manifold is a compact quotient of the tiling $\{n,m\}$, the order-$m$ $n$-gonal tiling. Let's say there are $E$ edges in total on this manifold. If we count the number of face-edge incidences, there are two per edge and $n$ per face, so we have $nF = 2E$, similarly if we count the number of vertex-edge incidences there are two per edge and $m$ per vertex, so $mV = 2E$. Together, we have $\chi(M) = V - E + F = (2/m - 1 + 2/n)E = (n/m - n/2 + 1)F$.
For the standard order-3 heptagonal tiling of the Klein quartic $K$, we have $m = 3$ and $n = 7$, so $\chi(K) = (2/3 - 1 + 2/7)E = -\frac 1{21}E = -\frac 16 F$. Since $\chi(K) = -4$, we find $F = 24$.
Another way you can see this is with the Gauss-Bonnet theorem applied to a surface of constant curvature $-1$: according to that, the area of a hyperbolic manifold $M$ is equal to $-2\pi\chi(M)$, and any geodesic polygon has area equal to its angular defect.
A regular Euclidean $n$-gon has interior angle sum $n \cdot (n-2)\pi/n$, and a $n$-gon in $\{n,m\}$ has interior angle sum $n \cdot 2\pi/m$, so the angular defect is equal to $(mn-2m-2n)\pi/m = -2\pi(n/m-n/2+1)$. Multiplying this by the number $F$ of faces, we obtain the total area $-2\pi(n/m-n/2+1)F$, but since we know this to be $-2\pi\chi(M)$ already we obtain the identity $\chi(M) = (n/m - n/2 + 1)F$ as before.
As for your question on how to find compact manifolds as quotients of hyperbolic reflection groups more generally, I've generally had success with the following method: First, via the above method determine the amount $F$ of fundamental domains (in your case triangles) you need to glue together to obtain the desired genus of manifold. Next, assign to each face (in your case edge) of the fundamental domain a permutation of $F$ elements such that each permutation decomposes into disjoint $2$-cycles and each product of two adjacent permutations decomposes into disjoint $2m$-cycles, where $\pi/m$ is the angle between those two corresponding faces (here edges). If such an assignment is possible, these permutations yield gluing instructions for your $F$ fundamental domains. For a given target genus there might actually be several nonisometric manifolds obtainable this way.
