How to show the Nirenberg problem is equivalent to solve $-\Delta u +1 =Ke^{2u}$?
What I try: because $g_{0_{ij}} = \langle \partial_i F , \partial _j F \rangle$, $F$ is position vector of $S^2$. Because $g=e^{2u}g_0$, so I can assume $\widetilde F=e^u F$. Then I to calculate the first and second fundamental form, but it is complex and different to (3).
Your $\tilde F$ does not produce the desired metric - you should get extra terms coming from the derivatives of $e^u$. In general you should not expect the new metric $g$ to come from an embedding of $S^2$ in $\mathbb R^3$, so you should use an intrinsic approach instead. Just start with $g = e^{2u} g_0$ and compute the geometry of $g$ in terms of $u$ and the geometry of $g_0$ - see e.g. this page for a bunch of examples. In dimension two you should be able to establish the formula
$$ K_g = e^{-2u} (K_{g_0} + \Delta_{g_0} u).$$
Since $g_0$ has constant curvature $K_{g_0} = 1$, this equation is equivalent to $(3)$ after changing the sign convention on $\Delta$.