### Solution:

#### Part (1) Proof:

Given the recursive formula for the sequence $\{a_{n}\}$:
$$a_{n+1}=Aa_{n}+B$$

We subtract $\frac{B}{1-A}$ from both sides of the equation:
$$a_{n+1}-\frac{B}{1-A}=Aa_{n}+B-\frac{B}{1-A}$$

Simplifying the right side:
$$=Aa_{n}+\frac{B(1-A)}{1-A}-\frac{B}{1-A}$$
$$=Aa_{n}-\frac{BA}{1-A}$$
$$=A\left(a_{n}-\frac{B}{1-A}\right)$$

Therefore, the sequence $\left\{a_n-\frac{B}{1-A}\right\}$ is a geometric sequence with common ratio $A$. This completes the proof.

#### Part (2) Solution:

Given $A=2$ and $B=-1$, we substitute these values into the formula for the sequence $\{a_{n}\}$:
$$a_{n+1}=2a_{n}-1$$

We know from part (1) that $\{a_n-\frac{B}{1-A}\}$ is a geometric sequence. Substituting $A=2$ and $B=-1$:
$$a_n-1=2^{n-1}(a_1-1)$$

Given $a_1=2$, we find $a_1-1=1$, so:
$$a_n-1=2^{n-1}$$
$$a_n=2^{n-1}+1$$

For the sequence $\{b_{n}\}$:
$$b_n=\frac{a_n-1}{a_na_{n+1}}=\frac{2^{n-1}}{(2^{n-1}+1)(2^n+1)}$$

This simplifies to:
$$b_n=\frac{1}{2^{n-1}+1}-\frac{1}{2^n+1}$$

To find the sum $T_n$ of the first $n$ terms of $\{b_{n}\}$:
$$T_n=\left(\frac{1}{2^0+1}-\frac{1}{2^1+1}\right)+\left(\frac{1}{2^1+1}-\frac{1}{2^2+1}\right)+\cdots+\left(\frac{1}{2^{n-1}+1}-\frac{1}{2^n+1}\right)$$

Notice the telescoping series, where most terms cancel out, leaving:
$$T_n=\frac{1}{2}-\frac{1}{2^n+1}$$

Simplifying the expression:
$$T_n=\frac{2^n-1}{2(2^n+1)}$$

Therefore, the sum of the first $n$ terms of the sequence $\{b_{n}\}$ is $\boxed{\frac{2^n-1}{2(2^n+1)}}$.