Let $ d $ be the degree of $ P(x) $. Then the degree of $ P(P(x)) $ is $ d^2 $. Hence, the degree of $ P(P(x)) + P(x) $ is $ d^2, $ and the degree of $ 6x $ is 1, so we must have $ d = 1 $. Accordingly, let $ P(x) = ax + b $. Then \[a(ax + b) + b + ax + b = 6x.\]Expanding, we get $ (a^2 + a) x + ab + 2b = 6x $. Comparing coefficients, we get \begin{align*} a^2 + a &= 6, \\ ab + 2b &= 0.\end{align*}From the first equation, $ a^2 + a - 6 = 0, $ which factors as $ (a - 2)(a + 3) = 0, $ so $ a = 2 $ or $ a = -3 $. From the second equation, $ (a + 2) b = 0 $. Since $ a $ cannot be $ -2, $ $ b = 0 $. Hence, $ P(x) = 2x $ or $ P(x) = -3x, $ and the sum of all possible values of $ P(10) $ is $ 20 + (-30) = \boxed{-10} $.