Absolute values in logarithms in a solution of differential equationHow have the moduli signs disappeared...

$g+kv$ need not be positive, but it will have the same sign as $g+ku$, because the solutions cannot cross the equilibrium at $v=-g/k$. Hence, the quotient inside the logarithm is positive.
Ideally, you would not arrive at $\frac{1}{k}(\ln |g+kv|-\ln |g+ku|)=-t$ at all; there is a cleaner way of solving this ODE. Namely, one goes from $\ln |g+kv|=-kt+C$ (with indefinite constant $C$) to $g+kv=A{e}^{-kt}$, where $A$ takes the place of $\pm e^C$

If $k$,$g$,$u$
are all positive, you need to have $v$ large enough (i.e., not too negative) to make $g+kv>0$. Maybe that results from some part that you haven't told us...