As can be seen from the name, it is a "local equilibrium" distribution function. In a non-relativistic context, I mean: $$f^0(t, \boldsymbol{x})= A(t,\boldsymbol{x})\,e^{-\frac{m}{2 K_B T(t, \boldsymbol{x})}\big(\boldsymbol{v}-\boldsymbol{u}(t, \boldsymbol{x})\big)^2}$$ where $A(t, \boldsymbol{x})$ is the normalization factor depending on local density and local temperature.
Note that in "global equilibrium", $T(t,\boldsymbol{x})=T=\text{constant}$ and $\boldsymbol{u}(t, \boldsymbol{x})=\boldsymbol{u}=\text{constant}$; obviously, $A$ will Also be a constant.
The surprising point is that this function doesn't solve the Boltzmann equation you wrote: $$\partial_t f^0(t,\boldsymbol{x})+\boldsymbol{v}.\boldsymbol{\nabla}f^0(t,\boldsymbol{x}) \ne 0$$ However, it turns out that the left hand side depends only on $\boldsymbol{\nabla}\boldsymbol{u}$ and $\boldsymbol{\nabla}T$. Here we need to include the "long wavelength limit assumption". In this limit, $f^{0}$ is the leading order solution in derivatives: $$\partial_t f^0(t,\boldsymbol{x})+\boldsymbol{v}.\boldsymbol{\nabla}f^0(t,\boldsymbol{x}) = \mathcal{O}(\boldsymbol{\nabla})$$ To find the sub-leading solution, taking $f=f^0+f^1$, we arrive at $$\partial_t f^0(t,\boldsymbol{x})+\boldsymbol{v}.\boldsymbol{\nabla}f^0(t,\boldsymbol{x}) = -\frac{f^1}{\tau}$$ which simply gives $f^1$. Similarly we can go through the derivative expansion to find higher order solutions $f^n$ with $n>1$.
Hope this helps.
P.S. See the kinetic theory lecture notes by David Tong: https://www.damtp.cam.ac.uk/user/tong/kintheory/kintheory.pdf.
What I mentioned above is more or less from this reference.