12. Given the sequence $\left\{a_{n}\right\}$, where $a_{1}=1, n \cdot a_{n+1}=2\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(n \in \mathbf{N}_{+}\right)$, the general term formula is
12. $a_{n}=n$.

From $n a_{n+1}=2\left(a_{1}+a_{2}+\cdots+a_{n}\right)$, and $(n-1) a_{n}=2\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)$, subtracting the two equations gives $n a_{n+1}=(n+1) a_{n}$, which means $\frac{a_{n+1}}{n+1}=\frac{a_{n}}{n}=\cdots=\frac{a_{1}}{1}=1$. Therefore, $a_{n}=n$.