How would you call a geometric object, which is topologically a planar graph, but its plane coordinates of the vertices and the curves of the edges matter?
Imagine the map of roads in a city. Not only the connectivity between road junctions are important, but also how the road actually goes from A to B.
I thought I can call it “geometric graph”, but this term likely means something else. I also consider “planar graph embedding”. But I worry that the “coordinate-awareness” of this object is not highlighted enough.
You're basically saying that you have a planar graph $G=(V,E)$ which means that $G$ admits an embedding in $\mathbb{C}$ (what this means is purposefully a bit vague here). However, you're caring about a particular embedding $\varphi$ which is a particular pair of injective maps $\varphi_1:V\to \mathbb{C}$ and $\varphi_2: E\to C([0,1])$ such that:
$i)$ $\varphi(e)$ is simple for every $e$.
$ii)$ As oriented edges, $\varphi_2((v,w);0)=\varphi_1(v)$ and $\varphi_2((v,w);1)=\varphi_1(w)$
$iii)$ For $e\neq f$, we have that $\varphi_2(e;[0,1])\cap \varphi_2(f;[0,1])\subseteq \{\varphi_2(e;0),\varphi_2(e;1)\}\cap \{\varphi_2(f;0),\varphi_2(f;1)\}$
At least that is a fairly reasonable definition of what a planar embedding of a graph is. The point of the matter is that you want to fix an embedding as opposed to just knowing, abstractly, that $G$ is planar.
It also seems to be relevant to you, since $C([0,1])$ carries a curve-length, this also makes $(V,E)$ a weighted graph by declaring the weight of $e$ to be the curve-length of $\varphi_2(e)$.