I found this explanation somewhere and wrote it down in my personal notes. I will explain with an example that I think exemplify why Riemann-Stieltjes will provide the wrong answer.
First, let's remember how we can define the Riemann-Stieltjes integral below
\begin{equation}\int_0^t Z(x)dZ(x)=\lim_{n\rightarrow\infty}\sum_{k=1}^n(Z(t_k)-Z(t_{k-1}))Z(t_{k-1})\tag{*}\end{equation}
where $0=t_1<t_2<...<t_n=t$, when $n\rightarrow\infty$, $(t_k-t_{k-1})$ goes to zero and $Z(x)$ is continuous and has bounded variation. The Ito integral can be defined in the same way (assuming $Z(t)$ to be any Brownian Path). So, in this elementar definition there is not really any difference, it is just that each is dealing with different kind of functions. But the integration rules will be different.
We can rewrite the terms in the sum as:
$$\sum_{k=1}^n(Z(t_k)-Z(t_{k-1}))Z(t_{k-1})=\frac{1}{2}Z(t)^2-\frac{1}{2}\sum_{k=1}^n(Z(t_k)-Z(t_{k-1}))^2$$
To exemplify, let's assume now that $Z(x)=x$. We know how to solve the integral in this case, this is $0.5t^2$, as in the first term of RHS above. What happens with this function is that we can ignore the second term above, since it will go to zero (try partioning $[0,t]$ with $t_k=kt/n$, for example). This comes from the fact that $Z(x)=x$ has bounded variation.
Why can't we simple do this if $Z(t)$ is a Brownian Path? The problem here is that $\frac{1}{2}\sum_{k=1}^n(Z(t_k)-Z(t_{k-1}))^2$ won't go to zero when $n\rightarrow \infty$. In fact,
$$\lim_{n\rightarrow\infty}\frac{1}{2}\sum_{k=1}^n(Z(t_k)-Z(t_{k-1}))^2=t$$
This is so because we are considering infinite realizations of a normal variable with variance $t_k-t_{k-1}$ as $n$ goes to infinity. The Brownian Path limit above does not go to zero because it fails to satisfy bounded variation.
That is why, even though Ito and Riemann-Stieltjes integration depart from the same definition (*) the results are very different. If $Z(x)$ is a Brownian Motion we get:
$$\int_0^t Z(x)dZ(x)=\frac{1}{2}Z(t)^2-\frac{1}{2}t$$
While if $Z(x)$ is a continuous function with bounded variation we get
$$\int_0^t Z(x)dZ(x)=\frac{1}{2}Z(t)^2$$