Let $L(M)$ be the space of smooth maps from $S^1$ to $M$.
Let $D(M)$ be the subspace of $L(M)$ consisting of immersions $f : S^1 \to M$ where $f$ admits a unique pair of points $p,q \in S^1$, $p \neq q$ with $f(p)=f(q)$.
My understanding of your question, is you want to know if $D(M) \subset L(M)$ is a submanifold.
I think it's not.
"Most" (*) of $D(M)$ has co-dimension $m-2$ in $L(M)$, where $dim(M)=m$. But the problem is $D(M)$ is a stratified space. $D(M)$ has a subspace where $f'(p)$ and $f'(q)$ are linearly dependent. Call this $D'(M)$. The reason for the asterix is that $D(M) \setminus D'(M)$ has co-dimension $m-2$ in $L(M)$. This is a genuine submanifold, and you get "charts" by doing the Vassiliev resolution-of-singularities, as $D(M) \setminus D'(M)$ is just immersions that have a single regular double point.
So the problem is getting charts around points of $D'(M)$. They don't exist. You can make the argument in general, but let's just take $M = \mathbb R^3$ just to make it somewhat concrete. Take a local picture that looks like the graph of the functions:
$$ t \longmapsto (t,0,0)$$
i.e. the x-axis
and
$$ t \longmapsto (t,t^2,0) $$
So this these are meant to be conjugate to two intervals of a map $f : S^1 \to \mathbb R^3$, describing a quadratic-type double point.
We consider a little neighbourhood of $f$ by doing a small perturbation to $f$ in $L(M)$.
$$ t \longmapsto (t,t^2+a, bx+c)$$
This is the original map when $a=b=c=0$. Its in $D'(M)$ only for that original map. For $a>0$ there are no singularities at all, but for $a<0$ there are the singularities corresponding to all parameters that satisfy
$$\pm b\sqrt{-a}+c=0$$
but the central singularity, where $b=c=0$ is one with two regular double points.
So this means that if $f \in D'(M)$, there are no manifold charts for $D(M)$ in $L(M)$ centered around $f$, since there are four "sheets" of $D(M) \setminus D'(M)$ incident to these points. $D(M)$ isn't locally flat about these points.
edit: Here is how you argue that $D(M) \setminus D'(M)$ is a submanifold of $L(M)$. Given a point $f \in D(M) \setminus D'(M)$ it is an immersion $S^1 \to M$ which is an embedding except for a single regular double point, meaning $f(p)=f(q)$ for $p,q \in S^1$, $p \neq q$, and $f'(p)$ and $f'(q)$ are linearly independent. The tangent space to $f$ is the sections of the pull-back $f^o(TM)$ bundle. The tangent space to $D(M) \setminus D'(M)$ at $f$ is given by certain constrained tangent vector fields along $f$. Thinking of $f^o(TM)$ as the tangent space to $L(M)$, it's the subspace of $f^o(TM)$ such that if $p,q \in S^1$ are the double points, then the vector over $p$ and the vector over $q$ are both tangent vectors in $T_{f(p)}M$, and we demand that the components of these vectors normal to $img(Df_p) \oplus img(Df_q)$ agree. By design, this is a $(m-2)$-dimensional constraint. The Vassiliev resolution of singularities idea -- pushing off the double-point -- is the complementary space to the tangent space of $D(M)\setminus D'(M)$ in $L(M)$. So this is a splitting of $T_f(L(M))$ into $T_f(D(M)\setminus D'(M)$ direct sum with an $(m-1)$-dimensional space corresponding to the deformations that remove the singularity. This also gives you your submanifold charts for $D(M)\setminus D'(M)$ at $f$.
Note: I meant to use standard notation for pull-backs but raising an asterix in math mode does strange things. So I use $f^o(bundle)$ instead.
Technically, what I've described in "edit" is a splitting of the tangent space to $L(M)$ at a point $f \in D(M) \setminus D'(M)$. The sub-manifold chart for $D(M) \setminus D'(M)$ is precisely the same chart as you use for $L(M)$ at $f$, but pre-composed with this splitting isomorphism. That this map is a submanifold chart for $D(M)\setminus D'(M)$ amounts to the observation that you can break this flow up into an isotopy that carries $img(Df_p) \oplus img(Df_q)$, together with a motion of the two arcs in this 2-dimensional space (which isn't an isotopy, it's just a 1-parameter family of immersions).
