In Theorem 2  of  [these notes](https://cmsa.fas.harvard.edu/wp-content/uploads/2022/03/immersions-revised2.pdf
), Ralph Cohen reformulates the main theorem of Hirsch-Smale theory merely in terms of normal bundles.
In particular,  he says that if $N, M$ are two manifolds, $\dim N< \dim M$ then two immersions
$f_1, f_2:N\to M$ are  regularly homotopic if and only if their normal bundles $\nu(f_1)$ and $\nu(f_2)$ are *isomorphic*.

>Why it is enough to have isomorphic normal bundle to have regularly homotopic immersions? In other words, how does the *if*  part of the statement follows from the classical formulation of the Hirsch-Smale theorem?

I recall that the celebrated Hirsch-Smale theorem states that if $\dim N< \dim M$ or $N$ open, then the tangential map gives a bijection between the connected components
$$\pi_0(Imm(N,M)) \to \pi_0(Mono(TN,TM))$$
where $Imm(N,M) $ is the set of immersions and $Mono(TN,TM)$ is the set of monomorphisms (bundle maps $TN\to TM$ injective fiberwise).

**My thoughts**

Now, assume that $\nu(f_1)\simeq \nu(f_2)$ are isomorphic to some $\nu\to N$.
In view of Hirsch-Smale's theorem, we would like to show that $df_1: TN\to TM$ and $df_2:TN\to TM$ are homotopic through monomorphisms. 
The hypothesis imply the existence of  two *isomorphisms*
$$F_i: TN\oplus \nu \to TM$$
for $i=1,2$, where $F_i|_{TN} = df_i$.
Clearly, if we show that $F_1$ and $F_2$ are homotopic through isomorphisms, then $df_1$ and $df_2$ are homotopic through monomorphisms so we are done. 
However I don't see any good reason why $F_1$ should be homotopic to $F_2$.
Notice that we only know that $\nu(f_1)\simeq \nu(f_2)$ but we have some freedom in choosing the embedding $\nu\to TM$, constructing different $F_1, F_2$. It is enough to show that any of these are homotopic.

**A special case**

Let's study a special case, when $N =\mathbb S^n$ and  $TM\simeq \varepsilon^m$ is trivial. Then we are given two trivializations  $F_1, F_2$ of $TN\oplus \nu$, and we would like to show that these are homotopic (through trivialization).
An (oriented) trivialization is a section of the bundle of oriented $n+v$-frames, $SO(TN\oplus \nu)\to N$, where $v= \mathrm{rank} (\nu)$. Which has fiber $SO(n+v)$.
The obstructions to construct an homotopy lie $H^i(\mathbb S^n, \pi_i(SO(n+v))$, thus the obstruction to homotoping $F_1$ to $F_2$ lies in $\pi_n((SO(n+v)))$ (provided the two are cooriented).
However the latter homotopy group is in general non-trivial.
As I said above, we have some freedom, i.e. we can change the trivialization $F_1:TN\oplus \nu \to \varepsilon^m$, by twisting it with  an automorphism of $Aut(\nu)$, thus to conclude it would be sufficient to show that $\pi_n(SO(v))$ acts transitively on $\pi_n(SO(n+v))$ where the action is induced by multiplication by $A\in SO(v)\subset SO(n+v)$ (the inclusion is $A\to \begin{bmatrix} 1 & 0\\ 0& A\end{bmatrix}$).