Given two immersions $f_i: M_i^m\to N_i^n $, $i=0,1$ we can always construct an immersion $f_0\# f_1 : M_0\#M_1\to N_0\#N_1$ restricting to $f_i$ over the image of the punctured $M_i$ in the connected sum $M_i\setminus \operatorname{int}\mathbb D^m\hookrightarrow M_0\#M_1$. Indeed the connected sum operation accounts to embedding two disks, removing their interior and gluing the two manifolds along their new spherical boundary component. On the other hand since $f_i$ are immersions locally they are represented by the inclusion $\mathbb D^m\to \mathbb D^m\times\{0\}\subset \mathbb D^n$; by using the disks provided by these charts we see that the $f_i$s match on (neighbourhoods of) the boundary to give $f_0\#f_1$.

Returning to your question, given $f:M^{m}\to \mathbb R^{2m}$, consider an immersion $g:\mathbb S^m\to \mathbb S^{2m}$ with a single self intersection at points $p_0, p_1\in \mathbb S^m$ (this can be constructed by hand) and take the connected sum $f\# g: M\#\mathbb S^m\to \mathbb R^{2m}\#\mathbb S^{2m}$ (of course the disks must be disjoint from ${p_0, p_1}$).

This can be done more generally for any target manifold of dimension $2m$, not just $\mathbb R^{2n}$.

Given two immersions $f_i: M_i^m\to N_i^n $, $i=0,1$ we can always construct an immersion $f_0\# f_1 : M_0\#M_1\to N_0\#N_1$ restricting to $f_i$ over the image of the punctured $M_i$ in the connected sum $M_i\setminus \operatorname{int}\mathbb D^m\hookrightarrow M_0\#M_1$. Indeed the connected sum operation accounts to embedding two disks, removing their interior and gluing the two manifolds along their new spherical boundary component. On the other hand since $f_i$ are immersions locally they are represented by the inclusion $\mathbb D^m\to \mathbb D^m\times\{0\}\subset \mathbb D^n$; by using the disks provided by these charts we see that the $f_i$s match on (neighbourhoods of) the boundary to give $f_0\#f_1$.

Returning to your question, given $f:M^{m}\to \mathbb R^{2m}$, consider an immersion $g:\mathbb S^m\to \mathbb S^{2m}$ with a single self intersection at points $p_0, p_1\in \mathbb S^m$ (this can be constructed by hand) and take the connected sum $f\# g: M\#\mathbb S^m\to \mathbb R^{2m}\#\mathbb S^{2m}$ (of course the disks must be disjoint from ${p_0, p_1}$). The same works for any $N^{2m}$ in place of $\mathbb R^{2m}$.

This argument informally tells you that we can avoid using a bump function and instead use nice gluing maps (identifications), so that the functions that you want to glue are equal to nice local models .

Given two immersions $f_i: M_i^m\to N_i^n $, $i=0,1$ we can always construct an immersion $f_0\# f_1 : M_0\#M_1\to N_0\#N_1$ restricting to $f_i$ over the image of $M_i\hookrightarrow M_0\#M_1$ in the connected sum. Indeed in a neighbourhood of a point $f_i$ behaves as an embedding $\mathbb D^m\to \mathbb D^n$ and by the uniqueness of such embedding up to isotopy we can find a diffeotopy $F:\mathbb D^m\times [0,1]\to \mathbb D^n\times [0,1] $ such that $F(\cdot, i) = f_i$ ($F(\cdot, t)$ doesn't have to send $\partial \mathbb D^m$ to $\partial \mathbb D^m$ for all $t$). Since $M_0\#M_1$ is constructed by gluing a tube $\mathbb D^m\times[0,1]$ to $M_0\bigsqcup M_1$ this gives $f_0\#f_1$.

Returning to your question, given $f:M^{m}\to \mathbb R^{2m}$, consider an immersion $g:\mathbb S^m\to \mathbb S^{2m}$ with a unique self intersection (this can be constructed by hand) and take the connected sum $f\# g: M\#\mathbb S^m\to \mathbb R^{2m}\#\mathbb S^{2m}$.

Given two immersions $f_i: M_i^m\to N_i^n $, $i=0,1$ we can always construct an immersion $f_0\# f_1 : M_0\#M_1\to N_0\#N_1$ restricting to $f_i$ over the image of the punctured $M_i$ in the connected sum $M_i\setminus \operatorname{int}\mathbb D^m\hookrightarrow M_0\#M_1$. Indeed the connected sum operation accounts to embedding two disks, removing their interior and gluing the two manifolds along their new spherical boundary component. On the other hand since $f_i$ are immersions locally they are represented by the inclusion $\mathbb D^m\to \mathbb D^m\times\{0\}\subset \mathbb D^n$; by using the disks provided by these charts we see that the $f_i$s match on (neighbourhoods of) the boundary to give $f_0\#f_1$.

Returning to your question, given $f:M^{m}\to \mathbb R^{2m}$, consider an immersion $g:\mathbb S^m\to \mathbb S^{2m}$ with a single self intersection at points $p_0, p_1\in \mathbb S^m$ (this can be constructed by hand) and take the connected sum $f\# g: M\#\mathbb S^m\to \mathbb R^{2m}\#\mathbb S^{2m}$ (of course the disks must be disjoint from ${p_0, p_1}$).