If $f$ and $g$ are in the same path component of $\operatorname{Imm}(M, N)$, then there is a path $p : [0, 1] \to \operatorname{Imm}(M, N)$ such that $p(0) = f$ and $p(1) = g$; i.e. $p$ is a path connecting $f$ and $g$. Now consider $H : M \times [0, 1] \to N$ given by $H(x, t) = p(t)(x)$, then $H(x, 0) = p(0)(x) = f(x)$ and $H(x, 1) = p(1)(x) = g(x)$ and for any $t \in [0, 1]$, $H(x, t) = p(t)(x)$ with $p(t) \in \operatorname{Imm}(M, N)$; also note that such a homotopy defines a path in $\operatorname{Imm}(M, N)$ between $f$ and $g$. So $f, g \in \operatorname{Imm}(M, N)$ are regularly homotopic if and only if they are homotopic via immersions (i.e. for every $t \in [0, 1]$, $H(\cdot, t)$ is an immersion) - this is usually taken as the definition of regularly homotopic. Note, this is a stronger condition than saying $f$ and $g$ are homotopic as the latter has no restrictions on the intermediate maps.
More generally, if $C(M, N)$ denotes the space of continuous maps from $M$ to $N$, and $\operatorname{Emb}(M, N)$ the space of embeddings of $M$ into $N$, then
- $f, g \in C(M, N)$ are homotopic if and only if they are in the same path-connected component of $C(M, N)$,
- $f, g \in \operatorname{Imm}(M, N)$ are regularly homotopic if and only if they are in the same path-connected component of $\operatorname{Imm}(M, N)$,
- $f, g \in \operatorname{Emb}(M, N)$ are isotopic if and only if they are in the same path-connected component of $\operatorname{Emb}(M, N)$.
As $\operatorname{Emb}(M, N) \subseteq \operatorname{Imm}(M, N) \subseteq C(M, N)$, one sees immediately that if $f, g \in \operatorname{Emb}(M, N)$ are isotopic, then they are regularly homotopic and homotopic; these implications can also be observed directly from the definitions.
The geometric difference between a regular homotopy and an isotopy is the same as the geometric difference between an immersion and an embedding. Every embedding is an immersion but not vice verca (i.e. $\operatorname{Emb}(M, N)$ is usually a proper subset of $\operatorname{Imm}(M, N)$). At some point in the regular homotopy, one of the intermediate maps will be an immersion but not an embedding; for example, whenever the image of the sphere has self-intersection. One of the explicit regular homotopies between $f$ and $-f$ was given by Morin; halfway through the regular homotopy, the image of the sphere is Morin's surface (image below from Wikipedia) which has self-intersection, so it cannot be the image of an embedding.
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The proof requires two steps, both steps being separate $h$-principles; one is Smale-Hirsch theorem as you said and the second is Gromov's $A$-directed embedding theorem.
By restricting your embedding $V \times \Bbb R \to W \times \Bbb R$ to $V \times 0$, you have an embedding $V \hookrightarrow W \times \Bbb R$ and as you concluded, you would like to position this embedded copy of $V$ such that restriction of the projection $W \times \Bbb R \to W$ gives an immersion $V \looparrowright W$.
Let $k = \dim V, l = \dim W$. Let $A \subset \mathrm{Grass}(k, W \times \Bbb R)$ be open subset of the Grassmannian of $k$-planes consisting of non-vertical planes, i.e., the planes which intersects the vertical lines $\{w\} \times \Bbb R$ transversely for all $w \in W$. You would like to homotope $V \hookrightarrow W \times \Bbb R$ to an embedding whose differential lands in $A$.
There is no obstruction at the level of differentials; given the tangent $k$-plane field along $V$ in $W \times \Bbb R$, one can always nudge the plane field to one which lands in $A$; this is because $k < l$, so there are at least two degrees of freedom for a ($k$-dim) tangent plane along $V$ in $W \times \Bbb R$ (which is $l+1$-dim) to move, since $(l+1) - k \geq 2$ so that we can simply rotate.
There is one more condition for the $h$-principle to go through; $A$ needs to be a complete subset of the Grassmannian, see Ch 4.6 in Eliashberg-Mishachev. This can be easily checked in the case of the above subset.
Given all of this, one can apply Gromov's theorem to homotope the embedding to be "$A$-directed", i.e., find a new embedding $V \hookrightarrow W \times \Bbb R$ whose differential lands in $A$. This concludes the second step in the proof of the exercise.
For a visually transparent proof of this corollary of the $A$-directed embedding theorem see Rourke-Sanderson, "The compression theorem I" (II and III are also beautiful papers which I highly recommend)
Best Answer
The space of maps $\operatorname{Mon}(TS^2,T\mathbb{R}^3)$ projects to the space of maps $\operatorname{Map}(S^2,\mathbb{R}^3)$ by forgetting the fiber maps (i.e. restricting to the zero sections). The fiber over a map $u$ is then the monomorphisms from $TS^2$ to $u^* T \mathbb{R}^3$. This is a locally trivial fibration with contractible base (by contracting to a constant map at $0$), hence a product. So it is homotopy equivalent to the fiber. To compute the fiber, take $u={\rm const}$. Get maps from $T S^2$ to trivial $\mathbb{R}^3$ bundle over $S^2$, which is the same as maps $T S^2$ to $\mathbb{R}^3$ monomorphic on the fibers.
This is the space of sections of a locally trivial fibration over $S^2$ with fiber the non-compact Stiefel manifold of all 2-frames in $\mathbb{R}^3$ (pick a local trivialization on $T S^2$ and see that on that trivializing chart you just need a 2-frame at over every point). This is associated fibration to $T S^2$, so is given by clutching function and since the clutching is twice the generator of $\pi_1 (V_{3,2})$, and so is trivial, the fibration is trivial. This means the space of sections is just maps $S^2$ to the fiber, which by Gram-Schmidt retracts to $V_{2,3}$. Hence $\pi_0$ is $\pi_2(V_{2,3})=0$, as wanted.
Maybe another way to replace/rephrase that last paragraph, is to note that $T S^2$ sits in the trival $\mathbb{R}^3$ bundle over $S^2$, as in $T S^2 \oplus \nu = {\rm trivial}$, and given orientations of everything (which we are implicitly assuming, actually), a monomorphism extends uniquely to a map of the $\mathbb{R}^3$-fiber to $\mathbb{R}^3$, where the vector in $\nu$ is the unit normal to the image of the original monomorphism. The space of such maps is a subset of 3 by 3 matrices, and a partial Gram-Schmidt says that the space of all invertible matrices retracts to it. The same argument in the $S^2$ family says the fiber is homotopy equivalent to the space of maps from $S^2$ to $GL(3)$, which is again by Gram-Schmidt homotopy equivalent to maps from $S^2$ to $SO(3)$. This gives same conclusion...
Also, the eversion is done in Eliashberg-Mashchev book on h-principle, in section 4.2.