The construction of the general Kummer surface, i.e., a surface with 16 ordinary double points in $\mathbb{P}^3$, is well-known, but there is a particular configuration about it I don't understand, and it can be found in Griffiths-Harris page 773:
Let $p_0$ be any of the double points of $S$, let $\tilde{S}$ be the blow-up of $S$ at $p_0$, and consider the map $r:\tilde{S} \rightarrow \mathbb{P}^2$ obtained by projection from $p_0$ onto a hyperplane. The generic hyperplane section $C_h$ of $S$ through $p_0$ is a plane quartic curve with one double point at $p_0$, its proper transform $\tilde{C_h} \subset \tilde{S}$ its desingularization. $\tilde{C_h}$ thus has genus $2$, and since $r$ expresses $\tilde{C_h}$ as a two-sheeted cover of its image $L = h \cap \mathbb{P}^2 \cong \mathbb{P}^1$, $r$ must be branched at exactly six points over $L$.
Most of this is clear to me, but there are three things I don't quite get. Firstly, why is $C_h$ a quartic curve, i.e., where did the degree $4$ come from? Next, how do we see that $\tilde{C_h}$ is a two-sheeted over of $L$? And finally, what's the difference between $h$ and $C_h$? Here I suppose $h$ is the hyperplane mentioned just before the first full stop.
$S$ is a quartic surface, hence $C_h$ is a quartic curve.
A general line through $p_0$ intersects $C_h$ in $p_0$ with multiplicity 2 (because this point is singular) and two more points. The intersection at $p_0$ disappears after the blowup, hence the degree of covering is 2.
$h$ is a hyperplane and $C_h$ is its intersection with $S$.