The goal is to define the ring structure on bordism. We will construct a spectrum $E$ by constructing a collection of pointed spaces $E_n$ with maps $E_n \to \Omega E_{n+1}$ which are equivalences.
Example 1. If $R$ is a ring, then $HR$ is a ring spectrum. How do we actually define $HR \wedge HR \to HR$? The former is represented by $$ (HR \wedge HR)_ {n+m} = (HR)_ n \wedge (HR)_ m, $$ and the homotopy group of the first thing is, by reduced Künneth, $$ \pi_\ast ((HR \wedge HR)_ {n+m}) = \begin{cases} R \otimes_\mathbb{Z} R & \ast = n+m \br 0 & \ast \lt n+m. \end{cases} $$ Then we have $$ H^{n+m}((HR)_ n \wedge (HR)_ m; R) = \Hom(R \otimes_\mathbb{Z} R, R) $$ and this has the multiplication element. So we get a map $(HR)_ n \wedge (HR)_ m \to (HR)_ {n+m}$ and this defines $HR \wedge HR \to HR$.
Bordism#
We define $$ MU(n) = \operatorname{Th}(EU(n) \to BU(n)) = DU(n) / SU(n). $$ Now there is a map $U(n) \to U(n+1)$ that includes in the first $n$ coordinates, inducing a map of pointed spaces $e_n \colon BU(n) \to BU(n+1)$. Then we have $$ e_n^\ast EU(n+1) = EU(n) \oplus \epsilon $$ where $\epsilon$ denotes the trivial complex line bundle. Then it is a general fact about the Thom construction that $$ \operatorname{Th}(EU(n) \oplus \epsilon) \cong \Sigma^2 \operatorname{Th}(EU(n)). $$ Using this, we get a map $$ \Sigma^2 \operatorname{Th}(EU(n)) \to \operatorname{Th}(EU(n+1)). $$ This can be used to define the spectrum $MU$.
Now how do we get a multiplication map? We have maps $$ BU(n) \times BU(m) \to BU(n+m) $$ and then pulling back the universal vector bundle just gives the direct sum of the universal vector bundles. Then we see that $$ \operatorname{Th}(EU(n)) \wedge \operatorname{Th}(EU(m)) \cong \operatorname{Th}(EU(n) \times EU(m)) \to \operatorname{Th}(EU(n+m)). $$ This defines a map $$ MU \wedge MU \to MU. $$
We can also construct the unit map by looking at $$ S^0 \cong \operatorname{Th}(EU(0)) \to MU, $$ inducing $\mathbb{S} \to MU$. This is going to be the unit.
The reason this is called bordism is the following. Define $\Omega_\ast^U$ the complex cobordism ring as $$ \Omega_\ast^U = \lbrace \text{stably almost complex manifolds} \rbrace / \lbrace \text{stably almost complex cobordisms} \rbrace. $$ Here, a stably almost complex manifold is an manifold $M$ together with an embedding $e \colon M \to \mathbb{R}^N$ with an almost complex structure on the normal bundle $\nu(e)$. This has a ring structure, with addition coming from disjoint union and multiplication coming from products.
Proposition 2. We have $\Omega_\ast^U \cong \pi_\ast MU$.
Now if we look at the normal bundle $\nu(e)$, and look at a tubular neighborhood, we get a map $$ S^N \to \operatorname{Th}(\nu(e)). $$ This $\nu(e)$ is classified by a map $M \to BU(N-n)$. Then we get a map $$ S^N \to \operatorname{Th}(EU(N-n)) = MU(N-n). $$ So this defines a ring homomorphism $$ \Omega_\ast^U \to \pi_\ast MU. $$ This turns out to be an isomorphism.