= Solution
Let $\gamma_{ij}$ denote the spatial <induced metric> and $\gamma=\det\gamma_{ij}$. Its conformally rescaled version $\widetilde\gamma_{ij}=\gamma^{-1/3}\gamma_{ij}$ has determinant one. The <Jacobi determinant derivative formula> consequently gives
$$
\gamma^{ij}\dot\gamma_{ij}=\frac{\dot\gamma}{\gamma},
\qquad \widetilde\gamma^{ij}\dot{\widetilde\gamma}_{ij}=0.
$$
Thus the determinant measures the volume-changing part of the spatial metric, while the unit-determinant part contains shape changes. <Metric compatibility> for the spatial <covariant derivative> also gives $\gamma^{ij}D_iN_j=D_iN^i$.
There is a sign inconsistency in the PDF. Contracting its defining expression for $K_{ij}$, with its threading convention $dx^i-N^idt$ and its future normal $n^\mu=N^{-1}(1,N^i)$, gives
$$
\boxed{K=-\frac1{2N}\left(\frac{\dot\gamma}{\gamma}+2D_iN^i\right),}
$$
not the displayed trace with a minus sign before the divergence. This is the <trace of extrinsic curvature with a negative shift>. For a concrete check, take $\gamma_{ij}=\delta_{ij}$, $N=1$ and $N^i=bx^i$. The defining formula gives $K_{ij}=-b\delta_{ij}$ and $K=-3b$, whereas the printed trace would give $+3b$. Reversing the shift convention can produce the other trace formula, but also changes the shift terms in $K_{ij}$ and the normal; the two conventions cannot be combined.
For the requested expansion interpretation, set the <shift vector> to zero. The sign discrepancy then disappears. A small coordinate volume has proper volume proportional to $\sqrt\gamma$, and proper time along its normal is $ds=Ndt$. Its expansion rate is
$$
\theta=\frac1N\partial_t\log\sqrt\gamma=-K.
$$
Therefore
$$
\boxed{H_{\rm local}=-\frac K3=\frac1{3N}\partial_t\log\sqrt\gamma
=\frac1N\partial_t\log(\gamma^{1/6}).}
$$
This is the <local volume Hubble parameter>, the mean of the three local directional expansion rates. In a homogeneous isotropic background $\gamma=a^6$, it reduces to $\dot a/(Na)$. The interpretation does not require the local expansion to be isotropic: the trace-free part of the <extrinsic curvature> can still describe shear.
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