Typically, when one derives 3-volumes in a metric, you need to define some sort of foliation for the metric in the timelike variable. Then, each leaf of your foliation is labeled by a function $\tau(t,x,y,z)$ such that $\tau$ is constant on each leaf. In your case, you want spacelike surfaces, so $\nabla_{a}\tau$ should be timelike. Define $\alpha^{2} = -\nabla_{a}\tau\nabla^{a}\tau$, and the unit vector $n_{a} = \alpha \nabla_{a}\tau$.
Then, the 3-metric is $\gamma_{ab} = g_{ab} + n_{a}n_{b}$.
If you do a coordinate change from t to $\tau$, then $\gamma$ is simply the 3x3 spacelike block of the metric. So yes, if you just choose the $t$ coordinate from your original metric as the time coordinate, the leaves are indeed all spatially flat.
But, if you choose, $t = \tau + G(x)$, then we have $dt = d\tau + G'(x)dx$, and the metric becomes:
$$ds^{2} = -(1 - (vf)^{2})d\tau^{2} - 2d\tau dx\left(G'(1-(vf)^{2}) + vf\right) + dx^{2}\left(1 - G'^{2}(1-(vf)^{2}) - 2vfG' \right) +dy^{2} + dz^{2} $$
Which certainly looks like a mess. But the operative point is that we are free to choose $G(x)$ to be anything we want. And in particular, we kill the diagonal term if we choose:
$$G(x) = \int dx\frac{-vf}{1-(vf)^{2}}$$
And our metric reduces to:
$$ds^{2} = -(1-(vf)^{2})d\tau^{2} + \left(1 + \frac{(vf)^{2}}{1-(vf)^{2}}\right)dx^{2} + dy^{2} + dz^{2}$$
And now, if we foliate on $\tau = $ Constant, it's clear that the 3-determinant of our leaves will be $1 + \frac{(vf)^{2}}{1-(vf)^{2}} = \frac{1}{1-(vf)^{2}}$
But note that this is an even function in $x -> -x$