I think the answer to the first 2 questions is yes. Most of the details are in the thesis of Ralph Herbert:

*Herbert, Ralph J.*, **Multiple points of immersed manifolds**, Mem. Am. Math. Soc. 250, 60 p. (1981). ZBL0493.57012

The important construction here is the *$r$-tuple point manifold* $\Delta_r(f)$ of the immersion $f:M^m\looparrowright N^n$, which for $r\ge2$ is defined as follows. Let $F(M,r)\subseteq M^{(r)}$ be the ordered configuration space of $r$-tuples $(x_1,\ldots , x_r)$ of points of $M$ such that $x_i\neq x_j$ whenever $i\neq j$; it is an open submanifold of $M^{(r)}$. Now consider the restriction of the $r$-th Cartesian power of $f$ to this configuration space, which will by abuse of notation will be denoted $f^{(r)}:F(M,r) \looparrowright N^{(r)}$. Generically, $f^{(r)}$ is transverse to the thin diagonal $d_r(N)= \{(n,\ldots , n)\}\subseteq N^{(r)}$ (see Herbert; also Gollubitsky and Guillemin, *Stable mappings and their singularities*, Chapter III, Corollary 3.3). Then $$\Delta_r(f) :=(f^{(r)})^{-1}(d_r(N))= \{(x_1,\ldots , x_r)\in F(M,r) \mid f(x_i)=f(x_j)\mbox{ for all }1\le i,j\le n\} $$
is a submanifold of $F(M,r)$ of codimension $rn-n$, therefore of dimension $rm-(rn-n)=n-r(n-m)$. If $M$ is a closed manifold, then so is $\Delta_r(f)$ (compactness is not obvious, it uses the fact that $f$ is locally an embedding). Note that $\Delta_r(f)$ carries smooth free actions of the symmetric group $\mathfrak{S}_r$ and the symmetric group $\mathfrak{S}_{r-1}$ which permute the last $r$ and $r-1$ coordinates, respectively. Let $M_r(f):=\Delta_r(f)/\mathfrak{S}_{r-1}$.

Now consider the map
$$
\mu_r(f): M_r(f)\to M,\qquad (x_1,[x_2,\ldots , x_r])\mapsto x_1
$$
given by projection onto the first co-ordinate. This $\mu_r(f)$ can be shown to be an immersion. Its image is the set of $x\in M$ such that $|f^{-1}f(x)|\ge r$.

Now restrict to the case $r=2$. The immersion $\mu_2(f):M_2(f)\looparrowright M$ is not an embedding when $f$ has triple points or higher. But if we remove the points of the domain where $\mu_2(f)$ fails to be injective, we get an injective immersion onto you $\Sigma_2$ (the "pure" double points), which in fact is an embedding. (A perhaps more convincing argument using general position which works for all $\Sigma_r$ is given on page 25 of Herbert.)

Thus $\Sigma_2\subseteq M$ is an embedded submanifold, and the involution $\phi:\Sigma_2\to \Sigma_2$ is smooth as it is conjugate to the restriction of invoution on $M_2(f)=\Delta_2(f)$ which permutes the factors.

For triple points and higher, I don't see any group action on $\Sigma_r$, since when $f^{-1}f(x)=\{x,x_2,\ldots , x_r\}$ there is no natural way to order the set $\{x_2,\ldots, x_r\}$.

exactly$n$. So $\Sigma_2 $ is disjoint from $\Sigma_3$ (triple points are not double points for me, sorry if this is the most used definition). In other words, just consider the double points for the sake of this question but don't assume the map does not have triple points. $\endgroup$