We can examine the homogeneous space by looking at either $\mathrm{SO}^+(3,1)$ or $\mathrm{SL}_2\mathbb{C}$. The first acts on the projectivized light cone of $\mathbb{R}^{3,1}$, called the celestial sphere (and which I'll denote $\mathcal{S}$) and the second on the complex projective line $\mathbb{CP}^1$, AKA the Riemann sphere $\widehat{\mathbb{C}}$. I will explain both $\mathcal{S}$ and $\mathbb{CP}^1$, how $\mathrm{SO}^+(3,1)$ and $\mathrm{SL}_2\mathbb{C}$ act on them, what the stabilizers are, how the actions are related, and conclude from the orbit-stabilizer that the homogeneous space is simply $S^2$.
Let's consider $\mathrm{SO}^+(3,1)$ first. The usual pseudo-Euclidean space $\mathbb{R}^{3,1}$ has pseudo-inner product
$$ \langle x,y\rangle = x_1y_1+x_2y_2+x_3y_3-x_4y_4 $$
The light cone is the set of all null vectors $v$ satisfying $\langle v,v\rangle=0$. That is, these vectors satisfy $v_1^2+v_2^2+v_3^2=v_4^2$, and if we projectivize (mod out by the scalar action of $\mathbb{R}^\times$), we get the space $\mathcal{S}$ of null lines. Any null line is represented by a unique null vector $v$ with $v_4=1,(v_1,v_2,v_3)\in S^2$. Thus the celestial sphere $\mathcal{S}$ is diffeomorphic to the usual round sphere $S^2$.
To understand the stabilizer of a null line, we will compare it with the stabilizer of a null vector. Using the previous pseudo-inner product and $v=[0~0~1~1]^T$, $ H=\mathrm{SO}(1)\times\mathrm{SO}(1)\times\mathrm{SO}(1,1)$ within the null line stabilizer $\mathrm{Stab}(\ell)$ (where $\ell=\mathbb{R}v$), $H$ consists of matrices of the form
$$ \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & \varepsilon\cosh s & \varepsilon\sinh s \\
0 & 0 & \varepsilon\sinh s & \varepsilon\cosh s
\end{bmatrix},
\quad \varepsilon=\pm1, $$
and has $v$ as an eigenvector with corresponding eigenvalue $\varepsilon\exp s$. Thus, it acts transitively on $\ell^\times=\mathbb{R}^\times v$, the nonzero elements of the null line. I recommend the instructive exercise of then showing $\mathrm{Stab}(\ell)$ is a knit product of $H$ and $\mathrm{Stab}(v)$, that is $\mathrm{Stab}(\ell)=H\mathrm{Stab}(v)$ and $H\cap\mathrm{Stab}(v)=1$. (We will see in a bit it is actually a semidirect product.) Trying to figure out the general form of an element of $\mathrm{Stab}(v)$, the algebra, to my mind, suggests an easier route might be to diagonalize $H$ in different coordinates...
Now let's use the quadratic form $x_1^2+x_2^2+x_3x_4$ with corresponding pseudo-inner product
$$ \langle x,y\rangle = x_1y_1+x_2y_2+\tfrac{1}{2}\big(x_3y_4+x_4y_3). $$
I got this from polarizing. Note if we write $x_3=\tilde{x}_3+\tilde{x}_4$, $x_4=\tilde{x}_3-\tilde{x}_4$, the quadratic form is $x_1^2+x_2^2+\tilde{x}_3^2-\tilde{x}_4^2$, so the signature is correct. Our null vector in the new coordinates is given by $v=[0~0~0~1]^T$, with corresponding $H$ consisting of
$$ \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & \varepsilon\exp(-s) & 0 \\
0 & 0 & 0 & \varepsilon\exp(s)
\end{bmatrix},
\quad \varepsilon=\pm1, $$
in which $v$ is an eigenvector with corresponding eigenvector $\varepsilon\exp(s)$. Again, $\mathrm{Stab}(\ell)$ will be a knit product of $H$ and $\mathrm{Stab}(v)$, so we will examine the general form of a matrix in $\mathrm{SO}^+(3,1)$ fixing $v$:
$$ T =
\begin{bmatrix}
\ast & \ast & \ast & 0 \\
\ast & \ast & \ast & 0 \\
\ast & \ast & \ast & 0 \\
\ast & \ast & \ast & 1
\end{bmatrix}. $$
The inner products of the columns should match the inner products of the standard basis elements. In particular, the first two columns are orthogonal to the fourth, and similarly the third column $z$ satisfies $\langle z,v\rangle=\langle e_3,e_4\rangle=1$ (of course, $v=e_4$), so we can narrow down the form a bit:
$$ T =
\begin{bmatrix}
a & b & x & 0 \\
c & d & y & 0 \\
0 & 0 & 1 & 0 \\
u & v & w & 1
\end{bmatrix}. $$
The inner products of the first two columns show $\big[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\big]\in\mathrm{O}(2)$, but cofactor expansion along the fourth column shows the determinant is $ad-bc=1$, so in fact $\big[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\big]\in\mathrm{SO}(2)$. Multiply by
$$\begin{bmatrix}
a & b & 0 & 0 \\
c & d & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}^{-1} $$
and rename variables so we have some matrix
$$ T =
\begin{bmatrix}
1 & 0 & x & 0 \\
0 & 1 & y & 0 \\
0 & 0 & 1 & 0 \\
u & v & w & 1
\end{bmatrix}. $$
The inner products of the first two columns with the third show $u=-2x,v=-2y$, and the inner product of the third with itself is $0$ so $w=-(x^2+y^2)$. So let's define the subsets
$$ K = \left\{
\begin{bmatrix}
\cos\theta & -\sin\theta & 0 & 0 \\
\sin\theta & \phantom{-}\cos\theta & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix} \mid \theta\in[0,2\pi)
\right\} $$
$$ A = \left\{
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & \varepsilon\exp(-s) & 0 \\
0 & 0 & 0 & \varepsilon\exp(s)
\end{bmatrix} \mid
s\in\mathbb{R}, \varepsilon=\pm1,
\right\} $$
$$ N = \left\{
\begin{bmatrix}
1 & 0 & x & 0 \\
0 & 1 & y & 0 \\
0 & 0 & 1 & 0 \\
-2x & -2y & -(x^2+y^2) & 1
\end{bmatrix} \mid
(x,y)\in\mathbb{R}^2
\right\} $$
This is the so-called Iwasawa decomposition. The letters stand for kompact (referring to the Lie group $K$), abelian and nilpotent (referring to the lie algebra $\mathfrak{n}$).
Observe $K\cong\mathrm{SO}(2)$ and $A\cong\mathrm{SO}(1,1)$ are subgroups, and $KA$ is an internal direct product. Moreover, $N\cong(\mathbb{R}^2,+)$ is a subgroup, and $KA$ normalizes it - exercise. Indeed, $\mathrm{Stab}(\ell)=(K\times A)\ltimes N$ is isomorphic to the similarity group $\mathrm{Sim}(2)$, with $K$ acting by rotations, $A$ acting by homotheties, $N$ by translations.
With the first coordinates, we may use the subgroup $\mathrm{SO}(3)\times\mathrm{SO}(1)\subseteq\mathrm{SO}^+(3,1)$ to show $\mathrm{SO}^+(3,1)$ acts transitively on the celestial sphere $\mathcal{S}$. With the second coordinates, we saw the stabilizer of a null line $\ell$ is an isomorphic copy of the similarity group $\mathrm{Sim}(2)$. By the orbit-stabilizer theorem, the homogeneous space $\mathrm{SL}_2\mathbb{C}/\mathrm{Sim}(2)$ is the celestial sphere $\mathcal{S}\simeq S^2$.
The group $\mathrm{GL}_2\mathbb{C}$ acts on the complex projective line $\mathbb{CP}^1$.
Any 1D subspace of $\mathbb{C}^2$ may be represented uniquely by a vector with second coordinate $1$, except the "$x$-axis" represented by $[\begin{smallmatrix}1\\0\end{smallmatrix}]$. We may identify $\mathbb{CP}^1$ with $\mathbb{C}\sqcup\{\infty\}$ via $[\begin{smallmatrix}w\\z\end{smallmatrix}]\leftrightarrow\frac{w}{z}$, adopting the convention $\frac{1}{0}:=\infty$. Then the action $[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}][\begin{smallmatrix}w\\z\end{smallmatrix}]=[\begin{smallmatrix}aw+bz\\cw+dz\end{smallmatrix}]$ can be turned into $[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}]\frac{w}{z}=\frac{aw+bz}{cw+dz}$, or in other words by Mobius transformations $[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}]\xi=\frac{a\xi+b}{c\xi+d}$, with $\xi=w/z$.
Any 1D subspace may be represented by a unit vector $v$, with Householder reflection $I-2vv^\dagger$. Since this reflection is invariant under the substitution $v\mapsto e^{i\theta}v$, it is uniquely determined by the 1D subspace. The set of all these reflections forms a 2D sphere in the space $\mathfrak{h}_2'\mathbb{C}$ of traceless hermitian matrices with the Frobenius norm aka Hilbert-Schmidt inner product. In fact, this is stereographic projection from $\mathbb{C}$, see here. Indeed, this is $\mathrm{SU}_2\mathbb{C}$-equivariant, respecting the metric, which exhibits the accidental isomorphism $\mathrm{SU}_2\mathbb{C}=\mathrm{Spin}(3)$. Also if we let $S^3$ be the unit sphere in $\mathbb{C}^2$, the composition $S^3\to\mathbb{CP}^1\to\mathfrak{h}_2'\mathbb{C}$ is the celebrated Hopf fibration.
The unit circle $S^1$ in $\mathbb{C}$ (identified with a subset of $\mathbb{CP}^1$) corresponds in $\mathbb{C}^2$ to the locus of points $(w,z)$ with $|w|=|z|$, or the null line of the sesquilinear form $|w|^2-|z|^2$ with matrix $\mathrm{diag}(1,-1)$. Indeed, we can show any circle $|z-a|=r$ or line (i.e. "generalized circle") in $\mathbb{C}$ corresponds to the null line of a sesquilinear form on $\mathbb{C}^2$, corresponding to a hermitian matrix of negative determinant unique up to real scalar multiple. And conversely! There is a right action of $\mathrm{SL}_2\mathbb{C}$ on $\mathfrak{h}_2\mathbb{C}$ given by $X\cdot A:=A^\dagger XA$, which preserves the determinant and extends the aforementioned action of $\mathrm{SU}_2\mathbb{C}$, now effectively acting on generalized circles of $\widehat{\mathbb{C}}$ wearing the disguise of hermitian matrices. The determinant $\det[\begin{smallmatrix}w+x & y+zi \\ y-zi & w-x\end{smallmatrix}]=w^2-x^2-y^2-z^2$ has signature $(1,3)$, exhibiting the accidental isomorphism $\mathrm{SL}_2\mathbb{C}=\mathrm{Spin}(1,3)$.
It is a textbook exercise to show $\mathrm{PSL}_2\mathbb{C}$ acts sharply $3$-transitively on the so-called Riemann sphere $\widehat{\mathbb{C}}=\mathbb{C}\sqcup\{\infty\}$ by Mobius transformations, via cross-ratio. So, in particular, $\mathrm{SL}_2\mathbb{C}$ acts transitively, and the stabilizer of $\infty$ is the Borel subgroup
$$ B=\left\{\begin{bmatrix} a & b \\ 0 & 1/a \end{bmatrix}\mid a\in\mathbb{C}^{\times},b\in\mathbb{C}\right\}. $$
This acts on $\mathbb{C}$ by $z\mapsto a^2z+b$, so there is a double covering $B\to\mathrm{Aff}_1\mathbb{C}$ with kernel $\pm I_2$. We can identify three major subgroups of this: $K$ of diagonal matrices with conjugate phasor (unit complex number) diagonals, $A$ of diagonal matrices with positive real reciprocal diagonals, and $N$ of unitriangular matrices. Just as before, $B=(K\times A)\ltimes N$, like $\mathrm{Sim}(2)$ earlier.
Since $\mathrm{SL}_2\mathbb{C}$ acts transitively on the Riemann sphere $\widehat{\mathbb{C}}$ with point-stabilizer $B$, the orbit-stabilizer theorem implies the homogeneous space $\mathrm{SL}_2\mathbb{C}/B$ is simply the Riemann sphere. Ultimately, we have a commutative diagram relating this with the $\mathrm{SO}^+(3,1)$ situation:
$$ \require{AMScd}
\begin{CD}
B @>>> \mathrm{SL}_2\mathbb{C} @>>> \widehat{\mathbb{C}} \\
@VVV @VVV @VVV \\
\mathrm{Sim}(2) @>>> \mathrm{SO}^+(3,1) @>>> \mathcal{S}
\end{CD} $$
The Riemann sphere $\widehat{\mathbb{C}}$ stereographically projects to the celestial sphere $\mathcal{S}$ within the space $\mathfrak{h}_2\mathbb{C}$ of hermitian matrices, which is the last vertical map. The first two vertical maps are double coverings sharing the same kernel $\{\pm I_2\}$, so this is a morphism between bundles.
One way to derive this result is by applying a more general result which allows one to integrate along fibers of a submersion:
Oriented Submersions
Let $M^m$, and $N^n$ be oriented smooth manifolds with $m>n$, and $\pi:M^m\to N^n$ be a smooth submersion. We can define adapted oriented coordinates as a set of oriented coordinate charts such that $\varphi$ has local representative equal to the projection onto the first $n$ coordinates $\pi(x^1,\cdots,x^m)=(x^1\cdots,x^n)$. There is a unique induced orientation on each fiber $\pi^{-1}(p)$ such that $(x^{n+1},\cdots,x^m)$ is an oriented chart on the fiber for any adapted oriented coordinates $x^i$.
Let $\alpha\in\Omega^nN$ and $\beta\in\Omega^{m-n}M$ be compactly supported differential forms. We can separate the following integral:
$$
\int_M\pi^*(\alpha)\wedge\beta=\int_Nf_\beta\alpha \\
\text{where}\ \ \ f_\beta(p):=\int_{\pi^{-1}(p)}\beta|_{\pi^{-1}(p)}
$$
To show this, it suffices to first show it holds in the case that $\beta$ is supported in a single set of adapted oriented coordinates with domain $(0,1)^m\mapsto(0,1)^n$, where it is a straightforward computation. The general case can then be obtained using a partition of unity.
Riemannian Submersions
Let $\pi:M^m\to N^n$ be a Riemannian submersion with $m>n$ ($M$ and $N$ now have metrics, but need not be oriented). Locally, we can find orthonormal coframes $E^1,\cdots,E^m$ on $U\subseteq M$ and $e^1,\cdots,e^n$ on $V(U)\subseteq N$ such that $\pi^*(e^i)=E^i$ for all $p\in U$ and $i\in[1,n]$. Note that $e^1\wedge\cdots\wedge e^n$ is a Riemann volume form on $N$, $E^1\wedge\cdots\wedge E^m$ is a Riemann volume form on $M$, and $E^{n+1}\wedge\cdots\wedge E^m$ restricts a Riemann volume form on each fiber $\pi^{-1}(q)\cap U$, Choosing a smooth $f:M\to \mathbb{R}$ supported in $U$, we can choose local orientations induced by these volume forms, we can apply the above result to the integral of $f$, yielding the following:
$$
\int_Mf\ dV=\int_N\hat{f}dV \\
\text{where}\ \ \ \hat{f}(q)=\int_{\pi^{-1}(q)}f\ dV
$$
Where $dV$ denotes the Riemannian measure on each manifold. Once again this result holds globally, provided $f$ is compactly supported, by a partition of unity argument.
Homogeneous Spaces
In your case, you have defined the metrics so that the quotient map $\pi:G\to X$ is a Riemannian submersion, and each fiber $\pi^{-1}(x)$ is isometric to $H$. Thus we have
$$
\operatorname{vol}(G)=\int_G1dV=\int_X fdV \\
\text{where}\ \ \ f(x):=\int_{\pi^{-1}(x)}1dV=\int_H1dV=\operatorname{vol}(H) \\
\implies \operatorname{vol}(G)=\int_X \operatorname{vol}(H)dV=\operatorname{vol}(X)\operatorname{vol}(H) \\
$$
Best Answer
I realized that the mistake is in the line "then 𝑓 should be proportional to the 2 dimensional volume". This is a tempting statement since we might naively believe that the level cylinder set should give us the density but it is blatantly false. In order to see this, look at an easier and clearer problem: just try to compute the volume of the 2 sphere assuming the analogous (false) statement and one will get the wrong answer of $\pi^2$ rather than the correct answer of $4 \pi$.