Geometric interpretation of the Pontryagin square The Pontryagin square (at the prime 2) is a certain cohomology operation
$$
\mathfrak P_2: H^q(X;\Bbb Z_2) \to H^{2q}(X;\Bbb Z_4)
$$
which has the property that its reduction mod 2 coincides with $x\mapsto x^2$. Furthermore,
If $x\in H^q(X;\Bbb Z_2)$ is the reduction of an integral class $y$, then $\mathfrak P_2(x)$
is the mod 4 reduction of $y^2$.
For a definition of $\mathfrak P_2$, see e.g.:
Thomas, E.: A generalization of the Pontrjagin square cohomology operation. Proc. Nat. Acad. Sci. U.S.A., 42 (1956), 266–269.
Suppose now that $X = M^{2q}$ is a closed smooth manifold of dimension $2q$.
Then $x\in H^q(X;\Bbb Z_2)$ is Poincaré dual to a class $x' \in H_q(X;\Bbb Z_2)$.
By Thom representability, $x'$ is represented by a map $f: Q^q\to M$ in which $Q$ is a (possibly unorientable) closed $q$-manifold (by taking the image of the fundamental class).
By transversality, we can even assume that $f$ is an immersion.
Question: Is there an interpretation of $\mathfrak P_2(x)$ as a geometric operation on 
$f$?
(Note: By the properties of the Pontryagin square, if $Q$ and $M$ are orientable, then 
then $\mathfrak P_2(x)$ is represented by the self-intersection of $f$ reduced mod 4.)
 A: The following geometric interpretation of $\mathfrak{P}_2$, due to Morita, is not exactly what you are looking for but maybe it can be interesting  as well. Anyway, it was too long to be a comment.
Assume $q=2k$, so that the Pontrjagin square is a map 
$$\mathfrak{P}_2 \colon H^{2k}(X, \mathbb{Z}_2) \longrightarrow \mathbb{Z}_4.$$ Set $$z:= \sum_{t \in H^{2k}(X, \mathbb{Z}_2)} e^{2 \pi i \mathfrak{P}_2(t)}.$$ Then $$\textrm{Arg}(z)= \frac{\sigma(M)}{8} \in \mathbb{Q}/\mathbb{Z},$$ where $\sigma(X)$ denotes the signature of $M$. 
For a proof, see Gauss Sums in Algebra and Topology, by Laurence R. Taylor.
A: This is just a guess, based on your comment about the case when $Q$ and $M$ are both oriented.
Claim: If $M$ is oriented and $Q$ is unoriented (and perhaps nonorientable), then the self-intersection of $Q$ is well defined in $\mathbb Z$ if $q$ is odd and well-defined in $\mathbb Z_4$ is $q$ is even.
Proof: Let $Q'$ be a parallel copy of $Q$ which intersects $Q$ transversely.  Choose local orientations of $Q$ and $Q'$ near their intersection points so that the local orientations of corresponding points of $Q$ qnd $Q'$ agree.  Define an intersection number (dependent on these choices) by summing the signs of the intersections.  There are two types of intersections of $Q$ and $Q'$: (a) those coming from the non-triviality of the normal bundle of $Q$, and (b) those coming from intersections of $Q$ with itself (if $Q$ if immersed but not embedded).  Changing a local orientation does not affect the contribution of type (a) intersection points to the total intersection, since both the $Q$ and $Q'$ local orientations change together.  Type $b$ intersections come in pairs.  If $q$ is odd then the signs of these pairs always cancel, so changing local orientation has no effect.  If $q$ is even then changing a local orientation flips the signs of both points, so the total intersection changes by $\pm 4$.
So perhaps in the case when $M$ is oriented the Pontryagin square is given by the above mod 4 intersection number, just as in the case when $Q$ is oriented.
