Proof:

(1) Given that $a_{n} + T_{n} = 1$, and the product of the first $n$ terms of the sequence $\{a_{n}\}$ is $T_{n}$, we can write the relationship for the $(n+1)^{th}$ term as follows:

- From $a_{n} + T_{n} = 1$, we have $T_{n+1} = a_{n+1} \cdot T_{n}$.
- Substituting $a_{n+1} = 1 - T_{n+1}$ into the equation, we get $T_{n+1} = (1 - T_{n+1}) \cdot T_{n}$.
- Rearranging, we find $\frac{T_{n+1}}{T_{n}} + T_{n+1} = 1$.
- Solving for $\frac{1}{T_{n}}$, we get $\frac{1}{T_{n}} + 1 = \frac{1}{T_{n+1}}$.
- Rearranging, $\frac{1}{T_{n+1}} - \frac{1}{T_{n}} = 1$.

Therefore, the sequence $\left\{\frac{1}{T_{n}}\right\}$ forms an arithmetic sequence with a common difference of $1$.

(2) Given $a_{n} + T_{n} = 1$, for $n=1$, we have $a_{1} + T_{1} = 1$. Since $T_{1} = a_{1}$, we find that $a_{1} = \frac{1}{2}$, and thus $\frac{1}{T_{1}} = 2$.

Knowing that $\left\{\frac{1}{T_{n}}\right\}$ is an arithmetic sequence with the first term $2$ and common difference $1$, we can express $\frac{1}{T_{n}}$ as $n + 1$. Therefore, $T_{n} = \frac{1}{n+1}$.

For $n \geq 2$, we find that $a_{n} = \frac{T_{n}}{T_{n-1}} = \frac{\frac{1}{n+1}}{\frac{1}{n}} = \frac{n}{n+1}$. This formula also holds for $a_{1} = \frac{1}{2}$, confirming that $a_{n} = \frac{n}{n+1}$.

Hence, we calculate the difference ratio as follows:
$$\frac{a_{n+1} - a_{n}}{a_{n}} = \frac{\frac{n+1}{n+2} - \frac{n}{n+1}}{\frac{n}{n+1}} = \frac{1}{n(n+2)} = \frac{1}{2}\left(\frac{1}{n} - \frac{1}{n+2}\right)$$

Summing these differences from $n=1$ to $n$ yields:
$$\sum_{n=1}^{n} \frac{1}{2}\left(\frac{1}{n} - \frac{1}{n+2}\right) = \frac{1}{2}\left(\frac{1}{1} + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right) = \frac{3}{4} - \frac{1}{2}\left(\frac{1}{n+1} + \frac{1}{n+2}\right)$$

Since $\frac{1}{n+1} + \frac{1}{n+2} > 0$, we conclude that:
$$\frac{3}{4} - \frac{1}{2}\left(\frac{1}{n+1} + \frac{1}{n+2}\right) < \frac{3}{4}$$

$\boxed{\therefore \frac{a_2 - a_1}{a_1} + \frac{a_3 - a_2}{a_2} + \ldots + \frac{a_{n+1} - a_n}{a_n} < \frac{3}{4}}$