Is there a direct geometric argument to show that two smooth manifolds that are topologically bordant are actually smoothly bordant? This is the question to be addressed.
The response below is a guide to a more satisfying situation....
In dimension three there is Likorish's argument using Dehn twists that oriented three manifolds bound. There is also Rochlin's geometric proof that in dimension four the complete bordism invariant is the signature circa 1950 [Oleg Viro at Stonybrook, his student, has these early references....] Rochlin was the student of Pontryagin. Thom suggested that he deserved the 1958 Fields Medal less than these predecessors.
Above that dimension, the only purely geometric step, to my knowledge, is the Pontryagin–Thom construction (1953 Thom, 1942 Pontryagin)reducing bordism questions to homotopy questions.
Serre's 1950 thesis opened the way to compute homotopy groups and the rest is the history alluded to in the question. Nevertheless.....
There is the theory of surgery or what was known as "constructive cobordisms" which one can learn from Milnor–Kervaire (Annals 1963)
or Browder's book "Simply connected surgery theory" or Novikov's papers from early 60's, or Wall's book on "Non simply connected surgery"
There is also obstruction theory as explained in Steenrod's book "Topology of fibre bundles" circa late 50's.
With these three tools in hand one can try to do something step by step.
In my Princeton thesis January 1966 "Triangulating homotopy equivalences" I tried this for the problem of constructing a PL homotopy bordism of a homotopy equivalence to a PL homeomorphism when it was impossible to compute the torsion aspects in the smooth case and when the torsion aspects of the difference between PL bundles and smooth bundles was also unknown.
By a stroke of luck the two unknown torsion groups canceled each other and an obstruction theory for the question with explicit coefficients in $0$ $\mathbb{Z}/2$ $0$ $\mathbb{Z}$ $0$ $\mathbb{Z}/2$ $0$ $\mathbb{Z}$.... emerged. [At the thesis defense Steenrod asked how these explicit obstructions obstructions could be calculated. The question was unexpected but a very good one. They were like any obstruction theory explicitly ill-determined but in this case the effective obstructions could be determined using bordism theory itself.]
Takeaway, there may be a tweaked version of the challenge being discussed which has a step by step construction which is geometric and constructive yet still obstructed but now by more explicit groups related to the signature and Kervaire–Arf invariant of surgery & one might do something that is part geometric and part algebraic but still rather concrete.
I would try to first lift the challenge to trying to lift a PL bordism to a smooth bordism because the relative homotopy group between topological and PL is one $\mathbb{Z}/2$ which is described by a signature divided by $8$ and reduced mod $2$.
Novikov's theorem/information expanded by et al comprises this information.
(BTW The sad news appeared today that Serge Novikov is very ill.)
Then directly attack lifting a PL bordism to a smooth bordism. Now the elementary geometric argument of Thom from Colloque Topologie Mexico City 1957 suffices to understand the rational Pontryagin classes by geometry. And BTW Novikov's argument is built on this argument of Thom and both are geometric plus a dollop of Serre which is easy.
Conclusion: There is a chance to have a geometric understanding of this question using just basic algebraic topology/geometric information about manifolds with a dash of serre algebraic topology/ algebra .
Dennis Sullivan Friday December 13th, 2019.
I think all elements are representable by honestly framed manifolds.
Let $M$ be a closed $d$-manifold with a stable framing, and consider the obstructions to destabilising a stable framing. Asumng $M$ is connected, which we can arrange by stably-framed surgery, there is a single obstruction, lying in $H^d(M ; \pi_d(SO/SO(d)))$.
If $d$ is even then $\pi_d(SO/SO(d)) = \mathbb{Z}$ and this obstruction may be identified with half the Euler characteristic of $M$. (As $M$ is stably framed, its top Stiefel--Whitney class vanishes and so its Euler characteristic is even.) We can change $M$ to $M \# S^p \times S^{2n-p}$ by doing a trivial surgery in a ball, and the stable framing extends over the trace of such a surgery. By taking $p$ to be 1 or 2 we can therefore change the Euler characteristic by $\mp 2$: thus we can change $M$ by stably framed cobordism until its Euler characteristic is 0, whence the stable framing destabilises to an actual framing.
If $d$ is odd then then $\pi_d(SO/SO(d)) = \mathbb{Z}/2$ and the obstruction is obscure to me (it is realised by the stable framing induced by $S^d \subset \mathbb{R}^{d+1}$, and is non-trivial even in Hopf invariant 1 dimensions where $S^d$ does admit a framing). I can't see an elementary argument for $d$ odd, but I think it is nontheless true by the following.
Let $d=2n+1$ with $d \geq 7$ (lower dimensions can be handled manually). Consider the manifold
$$W_g^{2n} = \#g S^n \times S^n.$$
This has a stable framing by viewing it as the boundary of a handlebody in $\mathbb{R}^{2n+1}$. By doing some trivial stably-framed surgeries as above (with $p=2,3$ say, to keep it simply-connected), we can change it by a cobordism to a manifold $X$ having an honest framing $\xi$. I wish to apply [Corollary 1.8 of Galatius, Randal-Williams, ``Homological stability for moduli spaces of high dimensional manifolds. II"], to $(X, \xi)$. There is a map
$$B\mathrm{Diff}^{fr}(X, \xi) \to \Omega^{\infty+2n} \mathbf{S}$$
given by a parameterised Pontrjagin--Thom construction. Now there is a step that I would have to think about carefully, but I think that the choices made can be arranged so that $(X,\xi)$ has genus $g$ in the sense of that paper, and so taking $g$ large enough the map above is an isomorphism on first homology. But this has the following consequence: any element $x \in \pi_{2n+1}(\mathbf{S})$ is represented by the total space of a fibre bundle
$$X \to E^{2n+1} \overset{\pi}\to S^1$$
with a framing of the vertical tangent bundle (and the Lie framing of $S^1$).
(Again, I'm sure there must be a more elementary way of seeing this.)
Best Answer
I sketch the proof of Buoncristiano and Hacon:
Let $M$ be a parallelizable manifold of dimension $m$. Let $N$ be $M \times M \setminus U$, where $U$ is a tubular neighbourhood of the diagonal (invariant under the natural involution on $M\times M$.) The involution on $N$ can be induced from the antipodal involution on the sphere $S^q$ for a sufficiently big $q$ (i.e., one may chose a $\mathbb Z/2$-equivariant embedding $N\hookrightarrow S^q$). The boundary of $N$ is $M \times S^{m-1}$, and the involution on the boundary can be induced from that on $S^{m-1}$.
So factorizing out by the involution the manifold $N$ we get a manifold $N'$, its map $f$ to $\mathbb{RP}^q$, and the boundary of $N'$ is mapped into $\mathbb{RP}^{m-1} \subset \mathbb{RP}^q$. Take an $\mathbb{RP}^{q-m+1}$ in $\mathbb{RP}^q$ that intersects $\mathbb{RP}^{m-1}$ in a single point. If both $f$ and its restriction to the boundary are transverse to this $\mathbb{RP}^{m-1}$ (this can be supposed) then $f^{-1}(\mathbb{RP}^{q-m+1})$ is a manifold with boundary $M$. Q.E.D.