Regular homotopy and the weak Whitney embedding theorem 
Two maps $f$ and $g\in$ $Imm(M,N)$ are regular homotopic if they are homotopic via immersions $h_t:M \to N$ and the derivatives $Th_t:TM
 \to TN$ of $h_t$ define a homotopy of bundle monomorphisms $$TM\times
 [0,1] \to TN, \quad (v,t)\mapsto Th_t(v). $$

For an immersion $f:M\to N$ between two closed manifolds $M^m$ and $N^n$ with $2m < n$, the weak Whitney embedding theorem ensures that we can approximate it arbitrarily close with an embedding.
How can I use this fact to ensure, that there exists a regular homotopy $h:M\times  [0,1] \to N$ between $f$ and an embedding $g:M\to N$?

Since the set of immersions is open in $C^\infty(M,N)$, a homotopy $h$ from $f$ to an arbirary map $g$ consists of immersions, if $|| h_t(x)-f(x)||< \epsilon$ for all $x\in M$ and for some $\epsilon$. Now for every map $g$ "close enough" to $f$, we can find an homotopy fullfilling this property (by Proposition $15.8.3$ of Tammo tom Diecks Algebraic Topology). Finally Whitneys embedding theorem ensures that there exists an embedding "close enough" to $f$. Thus $f$ and an embedding $g$ are homotopic via immersions.
But how can I certify that the deriveratives of those immersions $h_t:M \to N$ combine to a homotopy of bundle monomorphisms?
 A: Let me start by telling that there is nothing to explicitly check about the homotopy of bundle monomorphisms. If you have a homotopy $h$ such that every $h_t:M\xrightarrow{} N$ is an immersion, each $h_t$ will induce a monomorphism from $TM$ to $TN$. Since $h$ is smooth, so is the induced homotopy $H:TM\times [0,1] \xrightarrow{} TN$. And each $H_t$ is a monomorphism because $h_t$ are immersions.
A nicer (but more advanced) approach to see this would be to use Smale-Hirsch Theorem, which says $Imm(M,N)$ and the space of bundle monomorphisms from $TM$ to $TN$ are weakly homotopy equivalent. In particular, their zeroth homotopy 'groups' are isomorphic, so we have $$\pi_0\big(Imm(M,N)\big) \cong \pi_0\big(Mono(TM,TN)\big)$$
Note that two immersions lie in the same path connected component if and only if they are regularly homotopic. Similarly, two bundle monomorphisms are in the same path connected component if they are homotopic through monomorphisms. Actually, the map inducing this weak homotopy equivalence is given by $f\mapsto (f,df)$, where on the right hand side $f$ is the map between base spaces and $df$ is the map between total spaces. I believe this map, together with Smale-Hirsch Theorem, resolves your problem and the rest of your argument seems to work.
A little extra:
If you know $n\geq 2m+2$, you can actually prove the result for any immersion $f$ and embedding $g$. Since $M\times [0,1]$ is a $(m+1)$-dimensional manifold, the set of immersions rel boundary are dense in $C^\infty\big(M\times[0,1],N\big)$. Now take an immersion $H$ of $M\times[0,1]$ in $N$ such that $M\times\{0\}$ and $M\times\{1\}$ are mapped to the images of $f$ and $g$, respectively. It is known that immersions can be made self-transverse, so let us assume $H$ is self-transverse. Because of the dimensions, the self-intersections of the image of $H$ must consist of points, which are finitely many and isolated by compactness. Choose $\varepsilon>0$ such that all self-intersections on $H\big(M\times [t-\varepsilon,t+\varepsilon]\big)$ lie on the same $H(M\times \{t_0\})$. Now by appropriate bum functions, one can move these self-intersections to distinct $t$-levels, making each $H_t$ an immersion. Note that we can even assume $f$ is an embedding here.
