I'm not sure if the Doppler shift explains the twin paradox, but it does illustrate the difference between the two twins.
When twin 2 blasts off, she sees the signals from twin 1 Doppler shifted lower (they arrive less often that once every t seconds). The minute she turns around, she sees the signals from twin 1 Doppler shifted higher (they arrive more often than once every t seconds).
If twin 2 sends out N signals on the way out, and N on the way back, then she sends out a total of 2N signals. Because of the Doppler shifts, she receives [itex]N \frac{\sqrt{1-\frac{v}{c}}}{\sqrt{1+\frac{v}{c}}}[/itex] signals on the way out from twin 1, and receives [itex]N \frac{\sqrt{1+\frac{v}{c}}}{\sqrt{1-\frac{v}{c}}}[/itex] on the way back. So the total number she receives from twin 1 for the whole trip is:
[itex]N \frac{\sqrt{1-\frac{v}{c}}}{\sqrt{1+\frac{v}{c}}} + N \frac{\sqrt{1+\frac{v}{c}}}{\sqrt{1-\frac{v}{c}}} = 2N \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} > 2N[/itex].
So the traveling twin receives more signals than she sent, so she concludes that the stay-at-home twin aged more.
If you look at it from the point of view of the stay-at-home twin, it's not exactly the same. It's still true that he sees the signals from twin 2 Doppler-shifted lower during the outward journey, and Doppler-shifted higher during the return journey. But the difference is that the stay-at-home twin doesn't see the signals from the traveling twin until long after the traveling twin turns around. That's because, it takes time for those blue-shifted signals to reach Earth, and in that time, the traveling twin is also traveling toward Earth. So by the time the signals reach Earth, the traveling twin is almost home. So the stay-at-home twin doesn't see half red-shifted signals and half blue-shifted signals; he sees mostly red-shifted signals and then a very brief time of blue-shifted signals.