Doppler's effect According to the formula to calculate the apparent frequency heard by an observer, the frequency is independent of the distance of the source from the observer. This means however close the observer is from the source the frequency would be constant as far as the velocity is constant.  
Let's consider a source moving with a velocity $\ v$ towards an observer.Let its distance from the observer at an instant before it crosses the observer be $\ ds$. The frequency heard by the observer is $\nu_1$. When the source is exactly on the observer the frequency is the original frequency with which the waves were emitted, i.e., $\nu_o$. When the source crosses the observer and it is a distance $\ ds$ on the opposite side of the observer and the frequency now becomes $\nu_2$.   
My question is:

This seems a very abrupt/instantaneous process although in our daily life it doesn't happen so abruptly. The change in frequency is not so sudden. Why does that happen?

 A: Assuming you aren't going to crash head on into the object emitting the sound you need to pass it as some distance $b$:

You are zooming along at a constant velocity $v$, but your speed relative to the emitter is not $v$ but $dr/dt$, where:
$$ r^2=b^2+x^2 $$
If we differentiate this wrt time we get:
$$ 2r\frac{dr}{dt} = 2x\frac{dx}{dt} $$
And a bit of rearranging gives:
$$ \frac{dr}{dt} = \frac{x}{\sqrt{b^2+x^2}}v $$
So your speed relative to the emitter changes smoothly as you approach it then recede, and there is no abrupt change in the frequency. The only way to get a discontinuous change would be if $b=0$, but then you and the emitter would get to know each other more intimately than you probably want.
A: The abrupt change happens under the assumption that the observer is a point on the path of the source.
In "real life" there is always a distance between the observer and the path, so the component of the source's velocity that creates the Doppler effect changes.
The effective velocity is $\frac{V}{\mathrm{cos~\theta}}$  where $\theta(t)$ is the momentary angle between the source's path and the shortest path to the observer.

A: So in fact, both answers, Rennie's and Morag's are right, and equivalent. 
This is just to add a simple everyday view of the same effect.  
That is what you hear when a train with a whistle approaches you on a track at some distance from you and passes you by, with the track at a distance b at the closest approach, as per @Rennie's figure (except think of the train as the green dot moving along the x axis, and you being the still red point - it is all relative motion, so either you or the train moving will give you the same effect) 
You'd hear it louder at it approaches you but blue shifted (a higher pitch), with the higher pitch getting lower (yes, you'd hear the pitch, or freq,  diminishing), till it comes the closest to you, and the freq reaches $\nu_0$, and then you'd continue to hear the pitch getting lower (and the volume lower also) as it goes away from you and from the nearest point of approach. When it is coming towards you, the highest freq you hear is when it is at infinity (i.e., further away), and as it passes you by (the pitch keeps on continuously getting lower and lower. Unless b = 0 and then you would not care. Otherwise it is a blue shift as it approaches you, which because of b not being zero becomes its true freq at the closest approach, and then you'd hear the redshifted freq getting lower, asymptotically to $\nu_2$
