10. For any positive integers $m, n$, define the function $f(m, n)$ as follows:
(1) $f(1,1)=1$;
(2) $f(m+1, n)=f(m, n)+2(m+n)$;
(3) $f(m, n+1)=f(m, n)+2(m+n-1)$.
(1) Find the explicit formula for $f(m, n)$;
(2) Let $a_{n}=\frac{\sqrt{f(n, n)}}{2^{n-1}}\left(n \in \mathbf{N}^{*}\right), S_{n}$ be the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$, prove: $S_{n}<6$.
(1) From condition (2), we can obtain:
$$
\begin{array}{l} 
f(2,1)-f(1,1)=2 \times(1+1)=2 \times 2 \\
f(3,1)-f(2,1)=2 \times(2+1)=2 \times 3 \\
\cdots \cdots \\
f(m, 1)-f(m-1,1)=2 \times(m-1+1)=2 \times m
\end{array}
$$

Adding the above $m-1$ equations, we get $f(m, 1)-f(1,1)=2(2+3+\cdots+m)$, and since $f(1,1)=1$, we have
$$
f(m, 1)=2(2+3+\cdots+m)+1=m^{2}+m-1
$$

From condition (2), we can obtain:
$$
\begin{array}{l} 
f(m, 2)-f(m, 1)=2 \times(m+1-1)=2 m \\
f(m, 3)-f(m, 2)=2 \times(m+2-1)=2(m+1) \\
\cdots \cdots \\
f(m, n)-f(m, n-1)=2 \times(m+n-1-1)=2(m+n-2)
\end{array}
$$

Adding the above $n-1$ equations, we get $f(m, n)-f(m, 1)=2[m+(m+1)+\cdots+(m+n-2)]$, and since $f(m, 1)=m^{2}+m-1$, we have
$$
f(m, n)=2[m+(m+1)+\cdots+(m+n-2)]+m^{2}+m-1=m^{2}+2 m n+n^{2}-m-3 n+1
$$
(2) Since
$$
f(n, n)=n^{2}+2 n \cdot n+n^{2}-n-3 n+1=(2 n-1)^{2}
$$

We have
$$
a_{n}=\frac{\sqrt{f(n, n)}}{2^{n-1}}=(2 n-1) \cdot\left(\frac{1}{2}\right)^{n-1}
$$

Thus,
$$
\begin{array}{c}
S_{n}=1 \cdot\left(\frac{1}{2}\right)^{0}+3 \cdot\left(\frac{1}{2}\right)^{1}+\cdots+(2 n-1) \cdot\left(\frac{1}{2}\right)^{n-1} \\
\frac{1}{2} S_{n}=1 \cdot\left(\frac{1}{2}\right)^{1}+3 \cdot\left(\frac{1}{2}\right)^{2}+\cdots+(2 n-3) \cdot\left(\frac{1}{2}\right)^{n-1}+(2 n-1) \cdot\left(\frac{1}{2}\right)^{n}
\end{array}
$$

Subtracting the two equations, we get:
$$
\begin{aligned}
\frac{1}{2} S_{n} & =1+2 \cdot\left(\frac{1}{2}\right)^{1}+2 \cdot\left(\frac{1}{2}\right)^{2}+\cdots+2 \cdot\left(\frac{1}{2}\right)^{n-1}-(2 n-1) \cdot\left(\frac{1}{2}\right)^{n} \\
& =1+\frac{1-\left(\frac{1}{2}\right)^{n-1}}{1-\frac{1}{2}}-\frac{2 n-1}{2^{n}}=3-\frac{1}{2^{n-2}}-\frac{2 n-1}{2^{n}}
\end{aligned}
$$

Therefore, $S_{n}=6-\frac{2 n+3}{2^{n-1}}, n \in \mathbf{N}^{*}$, so $S_{n}<6$.