= Solution
Put $\rho=1/f$ and denote the flow generated by $fF$ by $\psi_t$. It follows the same oriented orbits as $\phi_s$, with the clock determined implicitly by
$$
\psi_t(x)=\phi_{s(t,x)}(x),\qquad
t=T(x,s):=\int_0^s\rho(\phi_r x)\,dr.
$$
Because $f$ is positive on a compact manifold, $0<f_{\min}\le f\le f_{\max}$ and $f_{\min}t\le s(t,x)\le f_{\max}t$ for $t\ge0$. Negative times obey the corresponding absolute-value comparison. This makes it possible to transfer exponential estimates once the correct transverse graphs are found.
For $v\in E^s(x)$ define
$$
\boxed{\lambda_s(x,v)=\frac1{\rho(x)}\int_0^\infty
d\rho_{\phi_r x}(D\phi_rv)\,dr}.
$$
The integrand is bounded by a constant times $e^{-\lambda r}\|v\|$, since $d\rho$ is bounded and $v$ is stable. The integral converges uniformly on bounded bundle vectors, so it is continuous and linear in $v$. Consider the graph
$$
\widetilde E^s(x)=\{v+\lambda_s(x,v)F(x):v\in E^s(x)\}.
$$
We verify its invariance and contraction directly; the old <Anosov stable bundle> need not itself remain invariant under a <positive time change of a smooth flow>.
For a fixed old time $s$, set
$$
I_s(x,v)=\int_0^s d\rho_{\phi_r x}(D\phi_rv)\,dr.
$$
Differentiation gives $D_xT(v)=I_s(x,v)$ and $D_xT(F)=\rho(\phi_sx)-\rho(x)$. Differentiating the implicit clock at fixed $t$, then differentiating $\phi_{s(t,x)}(x)$, therefore yields for $w=v+aF(x)$
$$
D\psi_t(w)=D\phi_s v+
\frac{a\rho(x)-I_s(x,v)}{\rho(\phi_sx)}F(\phi_sx).
$$
For $a=\lambda_s(x,v)$ this becomes
$$
D\psi_t(w)=D\phi_s v+
\frac1{\rho(\phi_sx)}\int_s^\infty d\rho_{\phi_r x}(D\phi_rv)\,dr\ F(\phi_sx).
$$
The tail coefficient equals $\lambda_s(\phi_sx,D\phi_s v)$ by the flow composition law, proving graph invariance. Both the first term and the tail have <norm> at most $C'e^{-\lambda s}\|v\|$. Projection onto $E^s$ in the original splitting is uniformly bounded, hence $\|v\|\le C''\|w\|$. The clock comparison now gives
$$
\|D\psi_t w\|\le C'''e^{-\lambda f_{\min}t}\|w\|,\qquad t\ge0.
$$
For the <Anosov unstable bundle>, use the convergent past integral
$$
\lambda_u(x,v)=-\frac1{\rho(x)}\int_{-\infty}^0
d\rho_{\phi_r x}(D\phi_rv)\,dr,\qquad v\in E^u(x).
$$
The same differentiation formula gives the tail $-\rho(\phi_sx)^{-1}\int_{-\infty}^s d\rho(D\phi_rv)\,dr$ for $s\le0$, proving invariance and exponential contraction in negative new time.
Finally, these two graphs and $\mathbb RF=\mathbb R(fF)$ form a direct splitting. Projecting onto $TN/\mathbb RF$ identifies the graphs with the old complementary stable and unstable subspaces, so they intersect neither each other nor the centre. We have established the continuous invariant splitting and both uniform estimates, and therefore \b[the flow of $fF$ is Anosov].
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