An urgent question

You could just apply Doppler Effect formula twice. One for the object receving the wave and the other for the object reflecting the wave.
Since the object is approaching the radar, the frequency received by the object is apparently higher, or the wavelength is apparently shorter.
(d-d')/d = u/c, where c is the speed of wave, and d' is the new wavelength
d-d' = ud/c
d' = d-ud/c

When the object reflects the wave, the motion of the object shortens the wavelength again. Hence, wavelength received by the receiver d'', is
(d' - d'')/d' = u/c
d' - d'' = ud'/c = (u/c)(d-ud/c) = ud/c - d(u/c)^2
i.e. d'' = d' - ud/c + d(u/c)^2 = d - ud/c - ud/c + d(u/c)^2
neglecting the term d(u/c)^2 because it is small compared with ud/c, as c is large when compared with u

hence, d'' = d - 2ud/c = d(1-2u/c)
when compared with the Doppler Effect formula: d'' = d(1-v/c)
we have, v = 2u
That is to say, the apparent wavelength got by the receiver is the same as if the object is moving with speed 2u towards it..

If you imagine it is the false image of a mirror,
when the mirrow moves at u, the false image moves at 2u.