EDIT: Let's try and turn this disaster into something with a moderate amount of workability by making some simplifying assumptions. I wouldn't recommend reading the previous version.
First, let's assume $X$ is a finite cell complex, coming from a filtration by subcomplexes $X^{(d)}$. (We'll allow ourselves the possibility of $X^{(d)}$ being formed from $X^{(d-1)}$ by attaching multiple cells of any necessary dimension.) The Thom spectrum functor of an infinite complex is (at the least, equivalent to) the colimit of the Thom spectra on its finite subcomplexes, and since our questions are about cellular filtrations we're not losing important details by restricting to finite objects.
Let's also assume for simplification that we have a topological group $G$ that acts by based maps on $S^n$, and a map $\alpha: X \to BG$ classifying a principal $G$-bundle $P \to X$. From this, we get a Thom space $X^\alpha = S^n \wedge_G P_+$. Let's assume for simplicity that we're talking about the associated virtual bundle of degree zero so that the Thom spectrum is the suspension spectrum $\Sigma^{-n} \Sigma^\infty X^\alpha$.
For each cell $D^k \to X$, choose a lift $D^k \to P$, which extends uniquely to a $G$-map $G \times D^k \to P$. These give $P$ a $G$-equivariant cell structure. If $P^{(d)}$ is the preimage of $X^{(d)}$, then $P^{(d)}/P^{(d-1)}$ is a wedge of copies of $G_+ \wedge S^{j_i}$. Every new cell of $P^{(d+1)}$ is attached via an equivariant boundary map $G \times S^{k-1} \to P^{(d)}$. On Thom spaces we get a filtration by $S^n \wedge_G P^{(d)}_+$, with the quotients in the filtration being wedges of $S^{n+k}$. In particular, the attaching maps between different layers are determined by $G$-equivariant maps $G_+ \wedge S^{k-1} \to \bigvee G_+ \wedge S^{j_i}$; taking the Thom space turns this into maps $S^{n+k-1} \to \bigvee S^{n+j_i}$.
Now take this whole picture and take suspension spectra.
We have a suspension spectrum $Y = \Sigma^\infty P_+$, which has an action of the group algebra $\mathbb{S}[G] = \Sigma^\infty G_+$. Taking suspension spectra preserves colimits, so that we can translate space-level identifications. We find that the suspension spectrum of $X$ is $\mathbb{S} \wedge_{\mathbb{S}[G]} Y$ where $G$ acts trivially on $\mathbb{S}$, while the Thom spectrum is $S^0 \wedge_{\mathbb{S}[G]} Y$. Here we view $S^0 = \Sigma^{-n} \Sigma^\infty S^n$ as being equivalent to $\mathbb{S}$, but inheriting its appropriate $G$-action.
We also have a filtration of $Y$ by submodules $Y^d = \Sigma^\infty_+ P^{(d)}_+$. The layers $Y^d / Y^{d-1}$ are wedges $\bigvee \mathbb{S}[G] \wedge S^{j_i}$. Let's use this filtration to build the "resolution"
$$
Y^0 \leftarrow \Sigma^{-1} Y^1/Y^0 \leftarrow \Sigma^{-2} Y^2/Y^0 \leftarrow \cdots.
$$
On homotopy groups, this gives a chain complex $C$ with $C_d = \pi_* (\Sigma^{-d} Y^d / Y^{d-1})$ of (graded) free modules over $\pi_* \mathbb{S}[G]$. It has the following properties.
It gives a spectral sequence with $E_2$-term $H_{**}(C)$ converging to $\pi_* Y$.
Smashing over $\mathbb{S}[G]$ with $\mathbb{S}$ tensors this chain complex over $\pi_* \mathbb{S}[G]$ with $\pi_* \mathbb{S}$. This chain complex comes from the cellular filtration of $\Sigma^\infty X_+$, and its homology is the $E_2$-term of the cellular spectral sequence for the homotopy of $\Sigma^\infty X_+$. (Standard techniques allow us to say there's a spectral sequence starting with $Tor^{\pi_* \mathbb{S}[G]}_{***}(\pi_* \mathbb{S}, H_{**}(C))$ and converging to the $E_2$-term.)
Smashing over $\mathbb{S}[G]$ with $S^0$ tensors this chain complex over $\pi_* \mathbb{S}[G]$ with $\pi_* S^0$ (in this case, $G$ acts nontrivially!). This chain complex comes from the cellular filtration of the Thom spectrum, and its homology is the $E_2$-term of the cellular spectral sequence for the homotopy of $\Sigma^\infty X_+$. (Again, there's a spectral sequence starting with $Tor^{\pi_* \mathbb{S}[G]}_{***}(\pi_* S^0, H_{**}(C))$ and converging to the $E_2$-term.)
The upshot of this is that, in terms of attaching maps, both the suspension spectrum of $X$ and the Thom spectrum of this degree-zero virtual bundle have cellular filtrations with the same cells. The relation between the adjacent attaching maps is this: they have common lifts from $\pi_* \mathbb{S}$ to $\pi_* \mathbb{S}[G]$, but one specialization has $G$ act trivially and one has $G$ act via its action on spheres.
Examples help.
If $G = O(1)$, then $\pi_* \mathbb{S}[G]$ is literally the group algebra over $\pi_* \mathbb{S}$ on $\mathbb{Z}/2$. The principal bundle over your space is a chain complex of free modules over this group algebra (and if $X = \mathbb{RP}^\infty$, it is a resolution of $\pi_* \mathbb{S}$). The two specializations make the generator ${-1}$ acts trivially and nontrivially respectively.
If $G = U(1)$, then $\pi_* \mathbb{S}[G]$ is even more interesting: it is a graded-commutative ring $\pi_* \mathbb{S}[d] / (d^2 = \eta d)$, where $|d| = 1$. The element $d$ comes from the canonical element in $\pi_1 U(1)$. The principal bundle again gives us a chain complex of free modules over this (and if $X = \mathbb{CP}^\infty$, it is the resolution of $\pi_* \mathbb{S}$ by free modules of rank $1$ given by alternately multiplying by $d$ and $d + \eta$). The two specializations make the generator act by $0$ and by $\eta$ respectively.
If $X = S^k$, then we can get into gory detail. We can use the cell decomposition with one zero-cell and one $k$-cell; the principal bundle is then formed by attaching $G \times D^k$ to $G$ along an equivariant map $G \times S^{k-1} \to G$ (so we need to be careful when $k=1$); without loss of generality we can make a choice sending the basepoint to the identity. This is literally an element in $\alpha \in \pi_{k-1}(G) = \pi_k BG$. Take suspension spectra and unwind: our chain complex is a two-term complex $\pi_* \mathbb{S}[G] \leftarrow \Sigma^{k-1} \pi_* \mathbb{S}[G]$. The map if $k \gt 1$ is given by (the image of) our element $\alpha$, while if $k = 1$ it is multiplication by the difference $1 - \alpha$. The two specializations make $\alpha$ go to zero (recovering the suspension of $S^k$) and to the appropriate image of $\pi_{k-1} G \to \pi_{k-1} Map(S^n,S^n) \to \pi_{k-1} \Omega^n S^n$ (recovering the cone on $\alpha$ or $1-\alpha$).
Secondary maps, like the $d_2$ in the spectral sequences I mentioned, are harder; suddenly you are trying to relate attaching maps for cells of two different objects. There may be some last bit of voodoo possible by taking these, which are basically Toda brackets, and trying to lift these to something equivariant. Seems like an interesting problem.
Best Answer
Thanks to a very helpful discussion with Clark Barwick in the homotopy theory chat, I think I now understand what's going on here. In particular, the ring spectrum structure on the sphere spectrum $\mathbb{S}$ does come from a monoidal structure on $\text{Bord}$, but I was confused about how to transport this monoidal structure from the category to the spectrum.
The monoidal structure can be thought of as coming from the universal property of $\text{Bord}$: since it's the free symmetric monoidal $\infty$-category with duals on a point, the $\infty$-category of symmetric monoidal functors $\text{Bord} \to \text{Bord}$ can canonically be identified with $\text{Bord}$ itself, and hence $\text{Bord}$ naturally acquires a monoidal structure, which I'll call $\circ$, coming from composition of functors $\text{Bord} \to \text{Bord}$. This is exactly analogous to how the free abelian group $\mathbb{Z}$ on a point canonically acquires a ring structure, and compatible under group completion with how the free spectrum on a point, namely $\mathbb{S}$, canonically acquires a ring spectrum structure. Here we need to know that symmetric monoidal $\infty$-categories are enriched over themselves, but this ought to be true by analogy both with the case of abelian groups and with the case of spectra.
This is admittedly an indirect description. It's hard to attempt a more direct description because the resulting monoidal structure isn't all that interesting on objects, and trying to describe what it does on morphisms is what got us into this mess in the first place.
So let me trudge on. In the comments I explained that I thought a monoidal structure
$$\text{Bord} \times \text{Bord} \to \text{Bord}$$
couldn't induce the usual multiplication map on $\mathbb{S}$ because monoidal structures take an $n$-morphism and an $n$-morphism and return another $n$-morphism: for example, the disjoint union does provide a monoidal structure of this form (in fact it's the monoidal structure figuring in the universal property), and the induced map
$$\pi_n(\mathbb{S}) \times \pi_n(\mathbb{S}) \to \pi_n(\mathbb{S})$$
is the usual abelian group structure on $\pi_n(\mathbb{S})$.
The problem with this story as applied to $\circ$ is that, with the natural symmetric monoidal structures on both sides, $\circ : \text{Bord} \times \text{Bord} \to \text{Bord}$ is not a symmetric monoidal functor, so it does not induce a map $\mathbb{S} \times \mathbb{S} \to \mathbb{S}$ of spectra. This issue shows up already at the level of abelian groups: with the natural abelian group structures on both sides, the multiplication map $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ is not a homomorphism of abelian groups.
As suggested by the analogy to abelian groups, $\circ$ is "bilinear": it preserves disjoint unions separately in both variables. So the fix is to use a suitable notion of "tensor product" of symmetric monoidal $\infty$-categories (suitable meaning in particular that on symmetric monoidal $\infty$-groupoids, thought of as connective spectra, it reproduces the smash product) and think of $\circ$ as instead providing a symmetric monoidal functor
$$\text{Bord} \otimes \text{Bord} \to \text{Bord}$$
which reproduces the usual ring spectrum structure $\mathbb{S} \otimes \mathbb{S} \to \mathbb{S}$ (here I am using $\otimes$ for the smash product as well). Now to see how we get a multiplication map
$$\pi_n(\mathbb{S}) \times \pi_m(\mathbb{S}) \to \pi_{n+m}(\mathbb{S})$$
it suffices to recall that the smash product of $S^n$ and $S^m$ is $S^{n+m}$.
So, one way to answer the conceptual question "how, in this situation, did we start with an $n$-morphism and an $m$-morphism and get an $n+m$-morphism" is that the universal property of $\text{Bord}$ is extremely general: it naturally acts by endomorphisms on an object in any symmetric monoidal $\infty$-category with duals whatsoever, including the "loop spaces" of $\text{Bord}$ itself! The analogous statement in stable homotopy is that the stable homotopy groups naturally give rise to operations on the homotopy groups of any spectrum whatsoever, including the shifts of the sphere spectrum itself.