The deRham map gives an arrow $\theta:H_{dR}(X) \to H(X)^*$, and the algebra structure on the right is induced from the coalgebra structue on $H(X)$ given by the Alexander-Whitney coproduct: $\Delta(\sigma) = \sum \sigma_i \otimes \sigma^i$ where $\sigma_i$ is the front $i$-face and $\sigma^i$ is the back $i$-face. Denote by $\langle -,-\rangle : H_{dR}(X)\otimes H(X)^*\to \mathbb R$ pairing that gives rise to $\theta$.
In terms of the pairing, you are correct this is the same as stating that, for any two forms $\omega,\eta$ of degee $p,q$ and any singular chain $\sigma$ of degree $p+q$, we have that
$$\tag{1}\langle \omega\wedge \eta,\sigma\rangle = \langle \omega,\sigma_p\rangle\langle \eta,\sigma^p\rangle$$
which is the same as $\langle \mu,-\rangle =\mu_\mathbb R\langle - \otimes -,\Delta\rangle$. You also note that there is a problem of writing simplices as products. There is a way of fixing this problem, which is (as Spivak does and in another context, Serre does), for example, done by working with singular cubes instead of singular chains.
Concretely, compute $H_*(X)$ using maps $\sigma : [0,1]^*\to X$. The differential is another, but the coproduct is simply given by restricting to coordinates, as in the case of simplices (though more terms appear). Concretely, if $\sigma : [0,1]^p\to M$ is a singular cube, the coproduct of $\sigma$ is given by a sum
$$\sum_{A\sqcup B = [p]}\sigma_A^0\otimes \sigma_B^1 $$
where $(-)_S^\varepsilon$ means you set the $S$-coordinates of the subset $S$ of $[p] = \{1,\ldots,p\}$ to $\varepsilon$ and keep the coordinates of the complement. See page 17 in Serre's thesis.
In this case, if $\sigma$ is a singular cube and $\omega = fdx_I$ is a form of the same degree $p$, it is straightforward to check that
$$\langle \omega,\sigma\rangle = \int_{[0,1]^p} f(\sigma) J_\sigma dx_1\cdots dx_p$$
and then $(1)$ is an immediate consequence of Fubini's theorem, as you wanted. There is a natural isomorphism between $H(X)$ computed in two different ways, either from cubical or simplicial singular chains, coming from a map from the cubical complex of $X$ to the singular complex of $X$ that needs to respect the coproduct (this is the nontrivial part of you question, I think, but you can find this map say in Serre's thesis), and this completes the proof that the deRham isomoprhism is a map of algebras.
Steenrod operations are an example. Let's work mod 2. The fact that the cup product is commutative on cohomology (and in a "coherent" way, consistent for all spaces) but not on cochains essentially means that at the cochain level, there is a chain homotopy between $\alpha \smile \beta$ and $\beta \smile \alpha$, or what I really mean is between the cup product $\smile$ and the composition $\smile \circ \tau$ of the cup product with the map $\tau$ that interchanges the two factors. Call this homotopy $\smile_1$. If $\alpha$ is a cocycle, then so is $\alpha \smile_1 \alpha$, so it induces an operation on mod 2 cohomology. You can in fact find $\smile_i$ products on cochains for each $i$, and each of them induces an operation on cohomology.
Edit: for each integer $n \geq 0$, there is a family of operations $\text{Sq}^n: H^j(X; \mathbb{F}_2) \to H^{j+n}(X; \mathbb{F}_2)$, one for each $j$. These are the Steenrod squares, and they are defined by reindexing the operators $x \mapsto x \smile_i x$. If $x \in H^n$, then $\text{Sq}^n(x)=x^2$, but also the Steenrod squares are compatible with suspension. So for example while you cannot tell $\Sigma \mathbb{C}P^2$ apart from $S^3 \vee S^5$ using the ring structure, you can tell them apart using Steenrod operations: since there is a nontrivial operation $\text{Sq}^2: H^2(\mathbb{C}P^2) \to H^4(\mathbb{C}P^2)$, there is a nontrivial operation $\text{Sq}^2: H^3(\Sigma \mathbb{C}P^2) \to H^5(\Sigma \mathbb{C}P^2)$ (all cohomology here with mod 2 coefficients).
Best Answer
$\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}$I'm not sure if this is an answer to the question, since it does refer to $H_{\ast}(M, \ZZ)$, but I think it sheds some interesting light on why the problem is hard.
We begin with a flawed proof attempt. Write $\delta$ for the diagonal map $M \to M \times M$, and $\pi_1$ and $\pi_2$ for the projections from $M \times M$ onto its first and second factor. Let $\rho$ be an integer $k+\ell$-cycle and let $\alpha$ and $\beta$ be a $k$-form and an $\ell$-form on $M$. Then $\int_{\rho} \alpha \wedge \beta = \int_{\delta(\sigma)} \pi_1^{\ast}(\alpha) \wedge \pi_2^{\ast}(\beta)$. Suppose $\rho$ were homologous in $M \times M$ to $\sum \sigma_i \times \tau_i$, for various cycles $\sigma_i$ and $\tau_i$ in $H_{\ast}(M)$, with $\dim \sigma_i + \dim \tau_i=k+\ell$. Then we would have $$\int_{\rho} \alpha \wedge \beta = \sum \int_{\sigma_i \times \tau_i} \pi_1^{\ast}(\alpha) \wedge \pi_2^{\ast}(\beta) =\sum_{(\dim \sigma_i, \dim \tau_i) = (k, \ell)} \int_{\sigma_i} \alpha \int_{\tau_i} \beta.$$ This would prove the result. (The integrals over terms where $\dim \sigma_i \neq k$ would drop out. If $\dim \sigma_i<k$, then $\pi_1^{\ast}(\alpha)|_{\sigma_i \times \tau_i}=0$ so $\int_{\sigma_i \times \tau_i} \pi_1^{\ast}(\alpha) \wedge \pi_2^{\ast}(\beta)=0$, and likewise if $\dim \tau_i < \ell$.)
Unfortunately, this need not be true. Saying that $\delta(\rho)$ should always be homologous to such a class is saying that $\delta_{\ast} H_m(M)$ should lie in the image of $\bigoplus_{i+j=m} H_i(M, \ZZ) \otimes H_j(M, \ZZ) \to H_{m}(M \times M, \ZZ)$. But the map $\bigoplus_{i+j=m} H_i(M, \ZZ) \otimes H_j(M, \ZZ) \to H_{m}(M \times M, \ZZ)$ need not be surjective -- the cokernel is $\bigoplus_{i+j=m-1} \mathrm{Tor}_1(H_i(M, \ZZ), H_j(M, \ZZ))$ by the Kunneth theorem. There is no reason that $\delta_{\ast}$ needs to land in the image of $\bigoplus H_i \otimes H_j$; I give an example where it doesn't in this Mathoverflow thread,
So we have to work harder. The good news is that the full statement of Kunneth's theorem includes the statement that $$0 \to \bigoplus_{i+j=m} H_i(M, \ZZ) \otimes H_j(M, \ZZ) \to H_{m}(M \times M, \ZZ) \to \bigoplus_{i+j=m-1} \mathrm{Tor}_1(H_i(M, \ZZ), H_j(M, \ZZ)) \to 0$$ is (noncanonically) split. Choose such a splitting $$\iota: \bigoplus_{i+j=m-1} \mathrm{Tor}_1(H_i(M, \ZZ), H_j(M, \ZZ)) \to H_m(M, \ZZ).$$ Then we can write $\delta(\rho) = \sum \sigma_i \times \tau_i + \iota(\sum \phi_j)$ for some $\phi_j$ in Tor groups. But the good news is that Tor groups are torsion, so an integer multiple of $\phi_j$ is homologous to zero. This means that $\int_{\iota(\phi_j)} \pi_1^{\ast}(\alpha) \wedge \pi_2^{\ast}(\beta)$ must be zero. So we still have $\int_{\rho} \alpha \wedge \beta = \sum \int_{\sigma_i \times \tau_i} \pi_1^{\ast}(\alpha) \wedge \pi_2^{\ast}(\beta)$ and now the proof concludes as before. $\square$.
What I find interesting about this argument is two things:
(1) We never had to build cup product in $H^{\ast}(M, \ZZ)$, or prove its compatability with de Rham cohomology, we just tried to shove cycles around naively.
(2) Proving Kunneth with the full statement that the sequence is split is not hard. But I have only ever seen it done in a very algebraic way, not by explicitly writing a cycle on $X \times Y$ as $\sum \sigma_i \times \tau_j + \phi$ for some torsion $\phi$. Moreover, there is no canonical choice of splitting. This may explain why no one seems to be able to prove this in a natural way.