= Solution
Use the given <Pestov identity for a geodesic flow> for the function $\varphi$ constructed above. Since $VX\varphi=-2H\varphi$, its left side is $-4\|H\varphi\|_{L^2(\mu)}^2$. Moving this to the right gives
$$
0=\|X\varphi\|_{L^2(\mu)}^2+5\|H\varphi\|_{L^2(\mu)}^2
+\int_{SM}(-K)(V\varphi)^2\,d\mu.
$$
All terms are nonnegative, and $-K$ is uniformly positive by compactness. The smooth derivatives therefore vanish pointwise:
$$
X\varphi=H\varphi=V\varphi=0.
$$
Since $X,H,V$ span the tangent bundle, $\varphi=c$ is constant, component by component if necessary.
On each circle fibre, $V^2u+u=c$ is the elementary periodic differential equation $u_{\vartheta\vartheta}+u=c$. Its solutions are
$$
u(x,\vartheta)=c+a(x)\cos\vartheta+b(x)\sin\vartheta.
$$
The first harmonics transform as a covector when the local oriented orthonormal frame changes. Their coefficients are smooth, either directly from the smooth fibre equation or from their sine/cosine integral formulas. Thus they define a smooth <one-form> $\eta$ on $M$ with $u(x,v)=c+\eta_x(v)$.
Along a geodesic, the velocity is parallel, so differentiating gives
$$
Xu(x,v)=\left.\frac d{dt}\eta_{\gamma(t)}(\dot\gamma(t))\right|_{t=0}
=(\nabla_v\eta)(v)=\langle\nabla_v\eta^\sharp,v\rangle.
$$
Set $Z=\eta^\sharp/2$. Then
$$
\boxed{\beta_x(v,v)=2\langle\nabla_vZ,v\rangle}.
$$
Equality initially holds on unit vectors, then on every vector by homogeneity. The <polarization identity> yields the tensor identity
$$
\beta(v,w)=\langle\nabla_vZ,w\rangle+\langle\nabla_wZ,v\rangle
=(\mathcal L_Zg)(v,w).
$$
This proves that $\beta$ is a <potential symmetric 2-tensor>, with the precise factor of two in the requested convention.
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