Okay, you have done this when you have 3 dimensions and I think you are saying you can find the vector equation of a line through 2 given points: the line through (1, 2, 2) and (2, 1, 5) has "direction vector" (2-1)i+ (1-2)j+ (5-2)k= i- j+ 3k so we can write the vector equation as r= (i+ 2j+ 2k)+ lambda(i- j+ 3k).
One problem you have is that the single equation y= 4x+ 3 is that a single equation in 3 dimensions describes a plane, not a line. So you must mean "in the xy-plane" which means that z= 0.
Now, one way to do that is to find two points! if x= 0, y= 3 so one point the line passes through is (0, 3, 0). If x= 1, y= 7 so another point is (1, 7, 0). Use exactly the same method as above to find the vector equation of the line through those two points. You should see that the component of the k vector is 0: the line is always in the xy-plane, of course.
One thing that may be confusing you is that the is no one vector equation. The parameter, lambda, has no "geometrical" meaning and can be chosen almost arbitrarily. When we have all the other coordinates written as functions of the other (here y is a function of x) we can always use that one coordinate as the parameter. That is, if y= f(x), we can write x= lambda, y= f(lambda) and have the vector equation r= lambda i+ f(lambda) j.