Usually the process of sphere eversion starts with a sphere whose normals are pointing outward, undergoes some spatial transformation without creasing or pinching (a homotopy), usually using a halfway-model like the tobacco-pouch, ends up as a sphere whose normals point inward. The sphere has been turned inside-out.
Is this equivalent to the following scenario ?
A smooth 3d unit vector field defined on a sphere, which initially points away from the sphere's center (like normals) is continuously and smoothly (without creases or any singularties) transformed into a vector field which points towards the sphere's center (like the inside-out normals).
Does the idea of sphere eversion translate to the described scenario (e.g. by using the normals of the surface during eversion rather than the positions) ? Or is it even the same with a different perspective ? And if so how could the latter be derived from the former ?
No, the scenarios are not equivalent, and no such transformation exists.
In fact, the outward normal vector field and the inward normal vector field are not even homotopic to each other through $\mathbb R^3$-valued nowhere zero vector fields on $S^2$, for the following reasons.
Associated to $\mathbb R^3$-valued nonwhere zero vector field $X$ on $S^2$ is the continuous self-map $f_X : S^2 \to S^2$ defined by $$f_X(p) = \frac{X(p)}{\|X(p)\|} $$ The function associated to the outward normal vector field is the identity map $I(p)=p$, which has degree $+1$. The function associated to the inward normal is the antipodal map $-I(p)=-p$, which has degree $-1$. But two homotopic nowhere zero vector fields give homotopic self-maps, and degree is a homotopy invariant, so the outward normal is not homotopic to the inward normal through nonzero $\mathbb R^3$-valued vector fields.