Since the remainder is $ 2x + 1 $ when $ f(x) $ is divided by $ (x - 1)^2, $ we can write \begin{align*} f(x) &= q(x) (x - 1)^2 + 2x + 1 \\ &= q(x) (x - 1)^2 + 2(x - 1) + 3.\end{align*}Then \[\frac{f(x) - 3}{x - 1} = q(x)(x - 1) + 2.\]Let \[g(x) = q(x)(x - 1) + 2.\]By the Remainder Theorem, $ f(3) = 15, $ so \[g(3) = \frac{f(3) - 3}{3 - 1} = \frac{15 - 3}{3 - 1} = 6.\]Also, $ g(1) = 2 $. Let $ ax + b $ be the remainder when $ g(x) $ is divided by $ (x - 1)(x - 3), $ so \[g(x) = q_1(x) (x - 1)(x - 3) + ax + b.\]Setting $ x = 1 $ and $ x = 3 $ gives \begin{align*} a + b &= g(1) = 2, \\ 3a + b &= g(3) = 6.\end{align*}Solving this system, we get $ a = 2 $ and $ b = 0, $ so \[g(x) = q_1(x)(x - 1)(x - 3) + 2x.\]Then \begin{align*} f(x) &= g(x) (x - 1) + 3 \\ &= [q_1(x) (x - 1)(x - 3) + 2x](x - 1) + 3 \\ &= q_1(x) (x - 1)^2 (x - 3) + 2x(x - 1) + 3 \\ &= q_1(x) (x - 1)^2 (x - 3) + 2x^2 - 2x + 3.\end{align*}This tells us that the remainder when $ f(x) $ is divided by $ (x - 3)(x - 1)^2 $ is $ \boxed{2x^2 - 2x + 3} $.