Why can't I define $\Delta u=(g(x)+\Delta x)-g(x)$?

Key concept: The confusion is very common in chain rule and differential notation. The key idea is that $\Delta u$ describes the change in the output caused by changing the input.

Step by step

The main reason

If $u=g(x)$, then $\Delta u$ must mean the change in $u$ caused by changing $x$ from $x$ to $x+\Delta x$.

So the correct definition is

$\Delta u=g(x+\Delta x)-g(x)$

because the new output is $g(x+\Delta x)$, not $g(x)+\Delta x$.

Why your idea does not work

Your proposal

$\Delta u=(g(x)+\Delta x)-g(x)$

simplifies to

$\Delta u=\Delta x$

That would mean the change in the output is always exactly the same as the change in the input. That is only true for special functions like $g(x)=x$, not for a general function.

What the instantaneous rate of change means

The derivative is based on the ratio

$\frac{\Delta u}{\Delta x}=rac{g(x+\Delta x)-g(x)}{\Delta x}$

and then taking the limit as $\Delta x\to 0$.

This measures how fast the function output changes because of the input change.

A quick example

If $g(x)=x^2$, then

$\Delta u=(x+\Delta x)^2-x^2=2x\Delta x+(\Delta x)^2$

This is clearly not equal to $\Delta x$.

Final idea

So $\Delta u$ cannot be defined as $(g(x)+\Delta x)-g(x)$, because that adds the input change directly to the function value instead of applying the function to the new input.

Pitfall alert

A common trap is mixing up the change in the input with the change in the output. $\Delta x$ belongs to the input variable, while $\Delta u$ must come from evaluating the function at the changed input.

Try different conditions

If $g(x)=x$, then your proposed expression happens to give the same result because the function is the identity. But for nonlinear functions such as $x^2$, $\sin x$, or $e^x$, the two definitions are completely different.

Further reading

chain rule, increment notation, derivative limit