My use of the technical language might be incorrect, so I'll try to explain it in simpler terms.
Let $G$ be a torus (or a sphere) in $\mathbb{R}^3$. I am looking for a function $f:G\rightarrow \mathbb{R}$ such that $d_G(x,y)=d_{\mathbb{R}}(f(x),f(y))$ where $d_{G}(x,y)$ is the length of the shortest path defined over the surface of the torus (or sphere) between points $x$ and $y$, and $d_{\mathbb{R}}(x,y)=|x-y|$.
I would image $f$ will be different for sphere and torus. There are related questions and resources such as this, and this that shows different geodesics over the torus, but none of them have an explicit form for $f$. Is there such explicit form?
I think the sphere is a little easier, since the geodesics are always given by arc lengths that depend on the angle between the two points. So, the problem becomes a matter of finding the angle between any two points in the sphere. However, here they consider differential equations, which to my naïve understanding seems overly complicated (it may be they are just being rigorous).
Overall, I'd appreciate if someone can share a resource or a closed form of $f$, or let me know in case getting such $f$ is not trivial.
I know there are many resources out there on differential geometry, but many use jargon and go deeper than what I want at the moment. With this question I am looking for answers/resources that keep things understandable even for those that don't know any differential geometry.
No such function exists. If $f$ exists as you have described then $f$ would be continuous and injective. To get a contradiction you can then apply the following general theorem: if $M$ is a 2-dimensional manifold then no continuous injective function $f : M \to \mathbb R$ exists. To prove this theorem, you would first apply the definition of a 2-manifold to obtain a continuous injection $g : D^2 \to M$, where $D^2 \subset \mathbb R^2$ is the closed unit disc, and then you would compose to get a continuous injection $f \circ g : D^2 \to \mathbb R$. Next, you can restrict to the unit circle $S^1 = \partial D^2$ to get a continuous injection $S^1 \mapsto \mathbb R$. But there does not exist any continuous injection $h : S^1 \to \mathbb R$: since $S^1$ is compact, $h$ has a maximum value $h(p)$ for some $p \in S^1$; you can then use the intermediate value theorem to find two points $s,t$ near $p$ such that $h(s)=h(t)$, contradicting injectivity.