(1) Since we have $a_1 = 2$ and $a_{n+1}=\frac{n+2}{n}S_n$ for $n=1,2,3,\ldots$,

We obtain $\frac{S_{n+1}}{n+1} = 2\cdot\frac{S_n}{n}$, with $\frac{S_1}{1} = 2 \neq 0$,

Therefore, the sequence $\left\{\frac{S_n}{n}\right\}$ is a geometric sequence with the first term $2$ and the common ratio $2$.

(2) From (1), we know that $\frac{S_n}{n}=2^n$,

Hence, $S_n = n \cdot 2^n$.

Now consider $T_n = 1\cdot 2 + 2\cdot 2^2 + 3\cdot 2^3 + \ldots + n\cdot 2^n$,

Let's compute $2T_n = 1\cdot 2^2 + 2\cdot 2^3 + 3\cdot 2^4 + \ldots + n\cdot 2^{n+1}$,

Subtracting these two equations, we get:
$$T_n - 2T_n = -(2 + 2^2 + 2^3 + \ldots + 2^n) + n \cdot 2^{n+1} = -\frac{2(1-2^n)}{1-2} + n \cdot 2^{n+1} = (1-n)2^{n+1} + 2$$

Solving for $T_n$ gives us $T_n = (n-1)2^{n+1} + 2$.

Hence, the sum of the first $n$ terms of the sequence $\{S_n\}$, denoted by $T_n$, is given by:
$$\boxed{T_n = (n-1)2^{n+1} + 2}$$