Baxter and Rennie say it better than me, so I will summarize them.

Suppose that $N_t$ is not stochastic and $f(.)$ is a smooth function then the Taylor expansion is $$ df(N_t) = f'(N_t)dN_t + \frac{1}{2}f''(N_t)(dN_t)^2 + \frac{1}{n!} f'''(N_t)(dN_t)^3 + \ldots $$ and the term $(dN_T)^2$ and higher terms are zero. Ito showed that this is not the case in the stochastic case. Suppose $W_t$ follows a Brownian motion and let $$ Z_{n, i} = \frac{W\left(\frac{ti}{n}\right) - W\left(\frac{ti}{n}\right)}{\sqrt{t/n}} $$ now consider $$ \int_0^t (dW_t)^2 \approx t \sum_{i=1}^n \frac{Z^2_{n,i}}{n}. $$ If $n \to \infty$ then $\sum_{i=1}^n \frac{Z^2_{n,i}}{n} \to 1$ and thus $\int_0^t (dW_t)^2 = t \neq 0$. Thus the second order term does not cancel (but higher order terms do) and we have an extra term in our derivative.

Baxter and Rennie say it better than me, so I will summarize them.

Suppose that $N_t$ is not stochastic and $f(.)$ is a smooth function then the Taylor expansion is $$ df(N_t) = f'(N_t)dN_t + \frac{1}{2}f''(N_t)(dN_t)^2 + \frac{1}{3!} f'''(N_t)(dN_t)^3 + \ldots $$ and the term $(dN_T)^2$ and higher terms are zero. Ito showed that this is not the case in the stochastic case. Suppose $W_t$ follows a Brownian motion and let $$ Z_{n, i} = \frac{W\left(\frac{ti}{n}\right) - W\left(\frac{ti}{n}\right)}{\sqrt{t/n}} $$ now consider $$ \int_0^t (dW_t)^2 \approx t \sum_{i=1}^n \frac{Z^2_{n,i}}{n}. $$ If $n \to \infty$ then $\sum_{i=1}^n \frac{Z^2_{n,i}}{n} \to 1$ and thus $\int_0^t (dW_t)^2 = t \neq 0$. Thus the second order term does not cancel (but higher order terms do) and we have an extra term in our derivative.