The relation $$ dt = (dW_t)^2 $$ expresses the idea that the square of increments of Brownian motion are nonzero even in the limit. As a result, when we Taylor expand a stochastic process the $dW_t^2$ terms cannot be discarded. For example, consider a random process $X_t$ defined by the stochastic differential equation $$ d X_t = \mu dt + \sigma dW_t. $$ If we discretise the stochastic differential equation then we have $$ \Delta X_t = \mu \Delta t + \sigma \sqrt{\Delta t} N(0, 1) $$ where $N$ is the normal distribution. Then, we have $$ (\Delta X_t)^2 = \sigma^2 N(0, 1)^2 \Delta t. $$ We can also see that the variance of the normal distribution implies $$ \text{E}\left[ N(0, 1)^2 \right] = 1. $$ So, when we perform a Taylor expansion on some function of $X_t$, $f(X_t)$, $$ \Delta f = \frac{\partial f}{\partial x} \Delta x + \frac{1}{2} \frac{\partial^2 f}{\partial x^2} (\Delta X_t)^2 + ... $$ in the limit $\Delta \rightarrow 0$, $(\Delta X_t)^2$ is proportional to $\Delta t$, and does not vanish. $$ $$ In contrast, for non-random functions when we perform a Taylor expansion $$ \Delta f = \frac{\partial f}{\partial x} \Delta x + \frac{1}{2} \frac{\partial^2 f}{\partial x^2} (\Delta x)^2 + ... $$ we discard all terms of higher order than $\Delta x$ in the limit $\Delta x \rightarrow 0$.