I give some easier example.
If beam of light travel down parallel with y axis in O frame. Then someone that at rest in O just see this light at rest respect to X axis, there is no motion along x direction. There is no velocity component of this light parallel to X axis. But what if beam of light travel at tilted path that make an angle with y axis. Then this beam will have velocity component that parallel with x axis and y axis, lets say it v_x and v_y that is less than c. So it is possible for some obserber to move with velocity = v_x and see this beam of light moving down with velocity = v_y.
I think it is not hard to understand.
No, it isn't hard to understand. You are wrong as you are using classical equations to describe a relativistic phenomenon.
Why don't you start with the simple 1D case. Once you grasp that, and realise you need to use relativistic formulas, you can move on to the 2D case.
The observer does not see the beam of light moving with a velocity of v_y as they are moving and their motion needs to be accounted for.
You are correct that the horizontal component of the velocity of the light will be 0 for this observer, but not the vertical velocity.
The proper formula would be this:
v_y*sqrt(1-v_x^2/c^2)/(1-v_x^2/c^2).
Multiplying the top and bottom by c^2 we get:
c*v_y*sqrt(c^2-v_x^2)/(c^2-v_x^2).
Noting that c^2=v_x^2+v_y^2, we get c^2-v_x^2=v_y^2.
Subbing that in we get:
c*v_y*sqrt(v_y^2)/v_y^2=c*v_y^2/v_y^2=c.
That means for the observer moving along, they will see the light travel vertically at a speed of c, not v_y.
There is no violation here.
Repeatedly ignoring this and continuing to use classical equations will not mean you are correct, it will mean you are wilfully ignorant.
If you have a rational object, then provide it, but stop appealing to classical equations that do not apply.
The problem is how to determined velocity component in term of velocity.
No it isn't.
If you try doing that in either reference there is no problem.
The problem is you pretending the 2 reference frames are the same and using classical mechanics.