Consider $f(t,x(t),y(t))$. Let $\Omega^*$ be the algebra over $\Bbb R$ that is generated by $dx,dy,dt$, and consider:
$$\Omega^*(\Bbb R^3)=\{C^\infty \text{ functions on }\Bbb R^3\}\otimes \Omega^*.$$
If I take the total derivative of $f(t,x(t),y(t)$ with respect to $t$, I obtain: $$\frac{d f}{dt}=\frac{\partial f}{\partial t}\frac{dt}{dt}+\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$$ $$\implies \frac{d f}{dt}=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$$
Then I can apparently multiply both sides of this by the differential $dt$. What does this actually mean mathematically?
It seems like a convoluted way of explaining exterior differentiation: $$df=\frac{\partial f}{\partial t} dt+ \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$$
But I might be missing the whole point.