Why do you need to use the chain rule in differentiation of ln?

It can be a bit strange to say that you need to use a specific formula or rule. I think it is more a question of you can use the formula.

So when can you use the chain rule? You can use the chain rule when you take the derivative of a composition of two functions. If $F(x) = f(g(x))$, then $F'(x) = f'(g(x))g'(x)$.

And if $F(x) = \ln(2x -1)$, then $F(x) = f(g(x))$ where $f(x) = \ln(x)$ and $g(x) = 2x-1$. And since $f'(x) = \frac{1}{x}$ and $g'(x) = 2$ you get $$ F'(x) = f'(g(x))g'(x) = \frac{1}{g(x)}g'(x) = \frac{1}{2x-1}2. $$

The theorem you are trying to use says

The derivative of $\ln x$ with respect to $x$ is $\frac{1}{x}$

So what about taking the derivative of $\ln(2x+1)$? Well, $\ln(2x+1)$ is not $\ln x$, so the theorem doesn't tell us anything about the derivative of $\ln(2x+1)$ with respect to $x$.

We invoke the chain rule so that we can rewrite the problem in a way involves finding the derivative of $\ln x$ along with some other stuff we know how to do.

Incidentally, some texts never write theorems like the one above, and always prefer to write theorems like

The derivative of $\ln f(x)$ with respect to $x$ is $f'(x) / f(x)$.

You need to use chain rule because it is a composition of functions: $f(x) = \ln(x)$ and $g(x) = 2x-1$, so we see $\ln(2x-1)$ as $f(g(x))$.