As @whuber said, there are so few possibilities here that a probability tree (comprehensive enumeration) is the simplest approach.
For the first draw at time $t = 1$, there are two possibilities: $R$ and $W$. $P(R) = 0.4$ and $P(W) = 0.6$. Here comes the tree:
IF a red was chosen at $t = 1$, THEN two reds are returned (one is added to the existing ten) and ball is drawn. Now at $t=2$ $P(R) = \frac{5}{11}$ and $P(W) = \frac{6}{11}$.
IF a white was chosen at $t = 1$, THEN three whites are returned (two are added to the existing ten) and ball is drawn. Now at $t=2$ $P(R) = \frac{4}{12}$ and $P(W) = \frac{8}{12}$.
So the probability of picking two different colored balls in the two draws would be red then white or white then red. Given the exhaustive list above, we can see that:
$$
\begin{aligned}
P(RW) = \frac{4}{10}\cdot\frac{6}{11} = \frac{12}{55}\\
P(WR) = \frac{6}{10}\cdot\frac{4}{12} = \frac{11}{55}\\
\rule{4cm}{0.3pt}\\
P(RR) = \frac{4}{10}\cdot\frac{5}{11} = \frac{10}{55}\\
P(WW) = \frac{6}{10}\cdot\frac{8}{12} =\frac{22}{55}
\end{aligned}
$$
So the probability of getting two different colors is:
$$
\frac{12}{55} + \frac{11}{55} = \mathbf{\frac{23}{55} \approx 41.8\%}.
$$
For completeness, it is good to see that the probability of getting two of the same color is $\frac{32}{55} \approx 58.2\%$ and that all the probabilities sum to 1, as they should.
