[nissanvietnam (OP)]

magine standing on a train platform, and throwing a ball at 30 km/h toward a train approaching at 50 km/h. The driver of the train sees the ball approaching at 80 km/h and then departing at 80 km/h after the ball bounces elastically off the front of the train. Because of the train's motion, however, that departure is at 130 km/h relative to the train platform; the ball has added twice the train's velocity to its own.

is it right or wrong?

[Bored chemist]

That's right.
In the same way, if you throw balls at the back of the train, they bounce back more slowly.

[syhprum]

Wrong ,the speeds relative the thrower simply add up although if you throw a ball at 30Kph towards a train receding at 50Kph no bounce occurs
"that departure is at 130 km/h relative to the train" This is incorrect the driver sees the ball departing at 30Kph relative the train

[Halc]

Wrong ,the speeds relative the thrower simply add up although if you throw a ball at 30Kph towards a train receding at 50Kph no bounce occurs

Indeed.  A 30 kph toss isn't going to catch up to a thing moving faster, but B-C didn't state the speed of the ball thrown at the rear, only implied that it was enough to hit the train.

"that departure is at 130 km/h relative to the train" This is incorrect

It is indeed incorrect, but nobody claimed that.

the driver sees the ball departing at 30Kph relative the train

The driver sees the shot to the front, so 80, as nissanvietnam points out.  An observer on the rear of the train would see a similarly thrown rear shot depart at 20 kph (50-30) without ever hitting.  This 30 figure doesn't match either of those, so not sure what you're talking about with it.

The driver of the train sees the ball approaching at 80 km/h and then departing at 80 km/h after the ball bounces elastically off the front of the train. Because of the train's motion, however, that departure is at 130 km/h relative to the train platform; the ball has added twice the train's velocity to its own.

is it right or wrong?

It's correct, but it makes some assumptions.
1, We're talking about low speeds here.  The calculations need to change if we ramp it up to relativistic speeds.  We're only doing Newtonian physics here, which works fine at the speeds you're using.
2, We assume the ball is quite small compared to the train and makes no significant change to the train speed during the collision.  For a big ball, the train slows down and the ball doesn't bounce back as much as if it were small. For a huge ball (like a planet), it is the train that bounces back.  One must perform a conservation of momentum calculation to get it right for all cases.

In all cases, the train guy sees the incoming ball and the receding ball moving at the same speed, regardless of relativity or mass of the ball.  In any fixed frame (platform say), the answer begins to differ if either of the assumptions above is not met.

[Janus]

This principle is what is taken advantage of when we use "gravity assists" by passing probes close to planets.  You don't have to throw the ball towards the train to get a Boost in speed.   If for instance, you were to toss the ball at 30 km/hr in the same direction as the trian's motion while in front of the train such that the train, moving at 50 km/hr catches up to a hits the ball "from behind", then the the ball and train collide with a relative speed of 20 km/hr, and the ball rebounds at 70 km/hr relative to the platform, for a gain of 40 km/hr.
With planetary gravity assists, we don't "bounce" the probe off the planet, but use the planet's gravity to change the probe's velocity.  In an ideal situation,  we would plan the probe's path around the planet to be a parabola, so that it makes a prefect 180 turn around the planet.   This isn't practical in real life, as in pretty much every case, the peirapis (closest approach to the planet's center) ends up being under the planet's surface.   Thus we have to settle for a hyperbolic trajectory and a less then maximum velocity exchange.   
We are also limited to using the " toss the ball in the same direction as the train's motion" method.   All the planets orbit the Sun in the same direction, and when we launch a probe, we take advantage of the Earth's orbital motion to get it up to the required velocity(with respect to the Sun).  This means the probe is going to be moving in the same direction as the planet when it get there.

[Bored chemist]

"that departure is at 130 km/h relative to the train" This is incorrect the driver sees the ball departing at 30Kph relative the train

OK, I can't swear to what the OP originally said but it does say "

hat departure is at 130 km/h relative to the train platform;

And the word "platform" there is critical to the meaning.