The main problem here seems to be that Mathematica is giving strange results when you ask it to take the derivative of the function Arg. As an example of this, suppose you take $\arg(e^{ix})$, where $x$ is a real number. You would expect that $\arg(e^{ix}) = x \; (\mathrm{mod} \; 2\pi)$, and therefore the derivative of this function should be a constant for most values of $x$. But this isn't what Mathematica does:
Plot[Evaluate[ReIm[D[Arg[Exp[I x]], x]]], {x, 0, 2}]
You do get the expected results if you insert a ComplexExpand, and if your expression is something that simplifies when you use ComplexExpand:
Plot[Evaluate[ReIm[D[ComplexExpand[Arg[Exp[I x]]], x]]], {x, 0, 2}]
However, ComplexExpand can't do much with an arbitrary InterpolatingFunction, so this method isn't suitable here. There may be some well-justified complex analysis reason why the first snippet above returns the given plot (it appears that the result is $i e^{2ix}$), but I'm not sure what it is just now. Alternately, it may be a bug.
To work around this, you can rewrite your equations so that they solve for the real and imaginary parts of $u$ separately, and then use the usual sorts of formulas to find the phase of $u$ and its derivatives:
eqns = {I D[u[t, x], t] + D[u[t, x], x, x] +
1/2 Abs[u[t, x]]^2 u[t, x] == 0, u[0, x] == 2 Sech[2 x]};
expandedeqns = Flatten[ComplexExpand[
ReIm[eqns /. {u[x_, t_] -> uR[x, t] + I uI[x, t],
Derivative[y__][u][x_, t_] ->
Derivative[y][uR][x, t] + I Derivative[y][uI][x, t]}]]]
sol = NDSolve[Evaluate[expandedeqns], {uR, uI}, {t, 0, 10}, {x, -10, 10}]
g[t_, x_] = ArcTan[uR[t, x], uI[t, x]] /. First[sol]
h[t_, x_] = D[g[t, x], x]
DensityPlot[h[t, x], {t, 0, 10}, {x, -10, 10}]
Finally, as noted by xzczd in the comments, you should not expect this answer to be correct unless you have fully specified the boundary conditions.