= Solution
Set $h=\det h_{ij}$. Contract the <extrinsic curvature with a negative shift> with $h^{ij}$. The <Jacobi determinant derivative formula> and metric compatibility give
$$
h^{ij}\dot h_{ij}=\partial_t\ln h=\frac{\dot h}{h},\qquad h^{ij}D_iN_j=D_iN^i.
$$
Consequently
$$
\boxed{K=-\frac{1}{2N}\left(\frac{\dot h}{h}+2D_iN^i\right).}
$$
The local volume element is $\sqrt h\,d^3x$, so for zero <shift vector> the proper-time derivative along the normal is $d/ds=N^{-1}\partial_t$ and
$$
\theta=\frac1N\partial_t\ln\sqrt h=-K.
$$
Here $\theta$ is the <expansion scalar>. Writing $a_{\mathrm{loc}}=h^{1/6}$ defines a local linear scale from this volume, whence the <local volume Hubble parameter> is
$$
\boxed{H_{\mathrm{loc}}=\frac{1}{a_{\mathrm{loc}}}\frac{da_{\mathrm{loc}}}{ds}=\frac{\dot a_{\mathrm{loc}}}{Na_{\mathrm{loc}}}=-\frac K3.}
$$
The factor of three converts volume expansion into linear expansion, and $N$ converts coordinate time into proper time. This agrees with the usual <Hubble parameter> in a homogeneous <FLRW metric>.
For a volume-shape decomposition the <unimodular spatial metric> is $\gamma_{ij}=a_{\mathrm{loc}}^{-2}h_{ij}$, with determinant one. If instead the printed positive power $a_{\mathrm{loc}}^2h_{ij}$ is taken literally, its determinant is $a_{\mathrm{loc}}^{12}$; it is a conformal rescaling but not the unimodular shape metric. The trace calculation does not need that rescaling.
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