Calculation with Ito processes, what is $ds \, dt$, $dW_t \, ds$ and $dW_s \, dW_t$?

The rules $$(dt)^2 = 0 \qquad dW_t \, dt = 0 \qquad (dW_t)^2 = dt \tag{1}$$ are heuristic rules to simplify calculations when applying Itô's formula. Mind that this is the only application; do not use them anywhere else.

In Itô's formula expressions of the form

$$\int_0^T f(X_t,Y_t) dX_t \, dY_t$$

pop up. Using $(1)$, we get

$$dX_t \, dY_t = (\mu_t dt + \sigma_t dW_t) (b_t \, dt + h_t \, dW_t) = \sigma_t h_t \, dt,$$

i.e.

$$\int_0^T f(X_t,Y_t) dX_t \, dY_t = \int_0^T f(X_t,Y_t) h_t \sigma_t \, dt.$$

As already mentioned above, this is the only application. When applying Itô's formula, you will never encounter

$$\int_0^T f(X_s,Y_t) \, dX_s \, dY_t$$

or similar things; therefore, it is enough to have $(1)$ and there is no need to ask for further rules.

Beware of the following mistake: Suppose that $$X_t = \int_0^t \mu_s \, ds \qquad \text{and} \qquad Y_t := \int_0^t b_s \, ds,$$ i.e. $dY_t = b_t \, dt$ and $dX_t = \mu_t \, dt$. Now it might be tempting to do the following calculation:

$$dX_t dY_t = \mu_t b_t (dt)^2 =0$$

and conclude that $X_t \cdot Y_t = 0$. This reasoning is not correct. To show that $X_t Y_t = 0$ we have to prove $d(X_t Y_t)=0$ and, in general, $d(X_t Y_t) \neq dX_t dY_t$. Just think of some examples where the processes can be calculated explicitly, e.g. set $\mu_t = b_t = t$. Then $$X_t = Y_t = \frac{t^2}{2}$$ and so $X_t \cdot Y_t = 0$ does, obviously, not hold true.

Remark: This question might be also of interest.