As can be seen here a sphere can be smoothly and continuously turned inside-out by a process called "sphere eversion". Let's call this scenario A.
On the other hand a 3d-unit-vector-field defined on a sphere can NOT be smoothly and continuously turned from pointing outward to pointing inward. Apparently this is so, because the function associated to the outward normal vector field has degree +1 while the inward pointing one has the degree -1. It follows that the two fields are not not homotopic. Let's call this scenario B.
What puzzles me is the following: I imagine that the vector field in B corresponds to the normals of the surface in scenario A (at the same fixed spherical coordinate). Or alternatively you could say, that the normal of the surface in scenario A maps to its equivalent fixed spherical coordinate in scenario B. Since scenario A start with a normal sphere, all vectors in B point outward. Now the sphere of A undergoes eversion, the surface bends and twists and so do the surface normals of A and therefore the vectors in B. When eversion is done, the sphere has been turned inside-out and so the surface normals in A are pointing inward and therefore the vectors in B point inward. But this can not be, because the impossible would have happend (inverting the vector field in B).
Where does my reasoning/analogy go wrong ?
A sphere eversion, by definition, is a regular homotopy $H : S^2 \times [0,1] \to \mathbb R^3$ from the identity map $H_0(p)=p$ to an orientation reversing diffeomorphism $H_1 : S^2 \to S^2$, where "regular" means simply that for each $t$ the map $H_t : S^2 \to \mathbb R^3$ is an immersion. For each $(p,t) \in S^2 \times [0,1]$, the oriented tangent plane $D_p H_t (T_p S^2)$ is therefore defined, and its unit normal vector $\nu(p,t)$ is defined. So from $H$ you get a continuous map $$NH : S^2 \times [0,1] \to \mathbb R^3 \times (\mathbb R^3 - \{0\}) $$ defined by $NH(p,t) = (H(p,t),\nu(p,t))$.
Now, one of the key observations in the study of regular homotopies is this.
As said in my answer to your earlier question, the identity map and the antipodal map are not homotopic as self-maps of $S^2$, and the "inward normal" vector and "outward normal vector" are not homotopic as maps $S^2 \mapsto \mathbb R^3 - \{0\}$.
However, consider that formula for $NH$. Restrict to $t=0$ and $t=1$, and you get the following two maps: $$NH_0(p) = (p,\nu_+(p)) \quad\text{and}\quad NH_1(p) = (-p,\nu_-(-p))$ $$ where $\nu_+$ means outward normal and $\nu_-$ means inward normal. Those are two smooth functions $$NH_0, NH_1 : S^2 \to \mathbb R^3 \times (\mathbb R^3 - \{0\}) $$ and those two functions $NH_0,NH_1$ are smoothly homotopic, because the second homotopy group $\pi_2(\mathbb R^3 \times (\mathbb R^3 - \{0\})$ is the trivial group, and a topological homotopy can be smoothed without much trouble.
The issue is, though, that one has no guarantee that $\mathbb R^3 - \{0\}$ coordinate is "normal to", or even "not parallel to", the tangent plane of the first coordinate map.
The mathematical issue underlying sphere eversion is therefore to start with a smooth homotopy from $NH_0$ to $NH_1$, and somehow use it to produce a smooth homotopy with the correct "normal vector" properties. What Smale realized is that there is a deeper homotopy invariant which is also trivial, and which implies, abstractly, the existence of the desired smooth homotopy. And once that was know, the existence of a sphere eversion follows. Armed with that knowledge, others later produced visualizable constructions.