### Solution:

#### Part (1): Proving that $\{\sqrt{S_{n}}\}$ forms an arithmetic sequence

Given that $a_{n+1} = S_{n+1} - S_{n}$ and $a_{n+1} = 1 + 2\sqrt{S_{n}}$, we can equate these two expressions to get:

$$S_{n+1} - S_{n} = 1 + 2\sqrt{S_{n}}$$

Rearranging this equation, we find:

$$S_{n+1} = S_{n} + 2\sqrt{S_{n}} + 1 = (\sqrt{S_{n}} + 1)^2$$

Taking the square root of both sides gives us:

$$\sqrt{S_{n+1}} = \sqrt{S_{n}} + 1$$

This implies that the difference between consecutive terms of $\{\sqrt{S_{n}}\}$ is constant, equal to $1$. Therefore, $\{\sqrt{S_{n}}\}$ forms an arithmetic sequence with the first term $1$ and common difference $1$.

#### Part (2): Finding the sum of the first $n$ terms of the sequence $\{c_{n}\}$

From part (1), we know that $\sqrt{S_{n}} = \sqrt{S_{1}} + (n-1) \cdot 1 = n$. Thus, we have:

$$c_{n} = (n + 1)2^n$$

The sum of the first $n$ terms of $\{c_{n}\}$, denoted as $T_{n}$, can be written as:

$$T_{n} = 2 \cdot 2 + 3 \cdot 2^2 + 4 \cdot 2^3 + \cdots + (n + 1) \cdot 2^n \tag{1}$$

Multiplying both sides of equation (1) by $2$ gives:

$$2T_{n} = 2 \cdot 2^2 + 3 \cdot 2^3 + 4 \cdot 2^4 + \cdots + (n + 1) \cdot 2^{n+1} \tag{2}$$

Subtracting equation (1) from equation (2) yields:

$$T_{n} = 2 \cdot 2^1 + 2^2 + 2^3 + \cdots + 2^n - (n + 1)2^{n+1}$$

Using the formula for the sum of a geometric series, we get:

$$T_{n} = 2 + \frac{2(1 - 2^n)}{1 - 2} - (n + 1)2^{n+1}$$

Simplifying the expression, we find:

$$T_{n} = n \cdot 2^{n+1}$$

Therefore, the sum of the first $n$ terms of the sequence $\{c_{n}\}$ is $T_{n} = n \cdot 2^{n+1}$. Encapsulating the final answer:

$$\boxed{T_{n} = n \cdot 2^{n+1}}$$