Problem

For a differentiable path $g(t)$, the ordinary chain rule gives

$$ f(g(T))-f(g(0))=\int_0^T f'(g(t))\,dg(t). $$

Why it fails for Brownian motion

Take

$$ f(x)=\frac12 x^2. $$

If the classical chain rule were valid for Brownian motion, then we would get

$$ \frac12 B_T^2=\int_0^T B_t\,dB_t. $$

But the right-hand side is an Ito integral, so its expectation is $0$, while

$$ \mathbb E\left[\frac12 B_T^2\right]=\frac T2>0. $$

Conclusion

Brownian motion needs a corrected chain rule, and the missing term is exactly what Ito formula provides.