I am trying to convince myself that a naïve definition of the Wall self intersection number should not work for odd-dimensional manifolds. Namely, let $X^{2n-1}$ be a smooth oriented closed manifold and let $f : S^n \to X$ be a map from a sphere. Also, let's assume that $X$ and $S^n$ are based and $f$ preserves base-points and $n \geq 2$. 


For now, suppose that $f$ is a smooth immersion with double point curves of intersection, and then by orienting the double point curves and attaching them to the basepoint along arcs contained in the image of $f$ to obtain $\mu(f) \in \mathbb{Z} \pi\_1(X)$. To remove the dependence on the orientations of the curves, I better mod out by $\{ g = g^{-1} : g \in \pi\_1(X)\}$. I would like for $\mu(f)$ to be invariant under changing $f$ by a homotopy, so I better also mod out by setting $1=0$. 


Is $\mu(f) \in \mathbb{Z}\pi\_1(X) /\langle\{ g = g^{-1} \}, 1=0\rangle$ an invariant of the based homotopy class of $f$?


I am guessing that the answer is no so ignoring everything above perhaps I should just ask: Is there a map $f : S^3 \to T^5$ that has exactly one double point curve in the image that is nontrivial in $\mathbb{Z}^5 = \pi\_1(T^5)$? 



Wall self-intersection invariant for odd-dimensional manifolds?
I doubt that what you are proposing as the receptacle for the obstruction is the correct abelian group. For one thing, you are not taking into account the involution on the canonical double cover of the double point manifold, i.e., 
 $\{(x,y)\in S^n\times S^n\setminus \Delta\_{S^n} |f(x) = f(y)\}$ (where we have assumed $f$ has been made self-transverse). This involution is a crucial aspect of the answer.


I think the answer to the question is going to be complicated.


Let me write a few words as to what might constitute a "non-naive" approach.


1) If $X$ is $2$-connected and $n > 2$, then there is no obstruction to deforming a map $f: S^n \to X$ to an embedding. This follows from the work of Haefliger, Hirsch, et. al.


So assume for the rest of this note that $X$ is not $2$-connected.


2) Consider the problem deforming, through regular homotopy, an immersion $f:S^n \to X^{2n-1}$ to an embedding, where $X$ is not assumed to be $2$-connected.


According to the work of Hatcher and Quinn (see also Corollary G my paper with Bruce Williams, "Homotopical Intersection Theory, II: Equivariance"), the obstruction, given by the double point manifold, lies in a certain twisted Bordism group
$$
\Omega\_0(E'(f,f)\_{h\Bbb Z\_2};\xi\_{\Bbb Z\_2}) \, ,
$$
where $E'(f,f)$ is the homotopy pullback of the diagram $S^n \to X \times X \overset{\Delta}\leftarrow X$ (the map $S^n \to X \times X$ is given by $x\mapsto (f(x),f(-x))$, and $\Delta$ is the diagonal). The space $E'(f,f)$ has a $\Bbb Z\_2$-action and $E'(f,f)\_{\Bbb Z\_2} = E\Bbb Z\_2 \times\_{\Bbb Z\_2} E'(f,f)$ is the Borel construction. Here $\xi$, defined over $E'(f,f)$, is a certain virtual vector bundle of rank $-1$, which is too technical to describe here.


Hence, we wish to identify the above bordism group as an abelian group.


Let us make a further simplifying assumption that the manifold $X$ is stably parallelized. In this case the virtual bundle is trivial. So the bordism group in this case is then $\pi\_0$ of the homotopy orbits of a certain action of $\Bbb Z\_2$ on the suspension spectrum $\Sigma^{-1} (E'(f,f)\_+)$. After choosing a basepoint in $X$, there is an evident equivariant map 
$\Omega X \to E(f,f)$ which is $(n-1)$-connected (where $\Bbb Z\_2$ acts by reversing loops). So the map
$\Sigma^{-1}(\Omega X\_+)\to
\Sigma^{-1} (E'(f,f))\_+$ is $(n-2)$-connected. 


Assuming that $n \ge 3$, the latter map will then be at least $1$-connected, and the obstruction be regarded as residing in $\pi\_0$ 
of the homotopy orbits of a certain action of $\Bbb Z\_2$ on the suspension spectrum $\Sigma^{-1} (\Omega X\_+)$. Here $\Bbb Z\_2$ is also acting on the suspension coordinate.


I don't know how to make this computation.


However, if we apply the "transfer" $\Omega\_0(E'(f,f)\_{h\Bbb Z\_2};\xi\_{\Bbb Z\_2}) \to \Omega\_0(E'(f,f);\xi)$ to this problem
(see my papers with Williams; this philosophically has the effect of thinking of the self-intersection question as an ordinary intersection question between two distinct manifolds and so we can then ignore the involution), then we are looking at the abelian group 
$$
\pi\_0(\Sigma^{-1} (\Omega X\_+)) = \pi\_{1}^{\text{st}}((\Omega X)\_+)
\cong \Bbb Z\_2 \oplus \pi\_1^{\text{st}}(\Omega X)\, .
$$
If we make yet another simplifying assumption that $X$ is $1$-connected, then
the displayed group is isomorphic to $\Bbb Z\_2 \oplus \pi\_2(X)$.
(But as noted above this group is solving a different question.)


3) There is the related question of deforming a map $f: S^n \to X$ to an embedding. The first step in solving the question is to find an immersion in that homotopy class. The obstructions can be studied using Smale-Hirsch theory; this essentially a question about find a section of a certain fiber bundle over $S^n$ whose fibers are Stiefel manifolds. 


Assuming the immersion exists we can look at all immersions in that homotopy class. Then we can employ the approach of (2) to each such immersion to create a "torsor" of indeterminacies. If we divide out by this indeterminacy, we get the answer. I will not detail this approach here, but a related problem of this kind was solved in a paper of mine, "Poincare complex diagonals."


**Addendum:** I just looked at the last section of Mark Grant's paper mentioned in the comments to the question. He is doing something similar to what I did at the end of (2) above, in that he is trying to avoid the problem of computing the involution.


