If the moon starts revolving the earth with relativistic speed, does this mean that moon's gravitational field on earth will increase and maybe even overpower earth's own field due to increased mass from the effect of relativity
$$ m = \frac{m_0}{\sqrt(1 - \frac{v^2}{c^2})} $$
Correct. The interaction depends on the speed of the moon in the suggested way for not too large Moon speeds.
Linearized gravity (Misner, Thorne, Wheeler, 'Gravitation', Ch 18.4) applies because the Earth has only weak gravity. Taking the Earth to be at rest, the interaction behaves like $$ \frac{GM}{r} \frac{m}{\sqrt{1-{v/c)^2}} \,.$$ With $v \approx 1 km/s$, the effect is only $1.1 \cdot 10^{-11}$. This approximation should hold up to quite high speed since the gravity of the Moon is so weak.
This follows from Eq. 18.15b in MTW. The second equality, '= Newton potential'), in this reference is only valid for low velocity.
G: gravitational constant
M: Earth mass
m: Moon mass
r: Earth - Moon distance
v: Moon velocity
