Are perpendicular components special in vectors? We can split a vector (velocity/displacement vector) along any two directions as long as the resultant of the oblique components of the vector is same as my original vector. Similarly if we have to add two vectors we can split them along any two axes and add components along these axes . The result will be the same either way.
Is there any good reason that we split vectors into mutually perpendicular vectors ? (except for that its easier to compute ).Is there any deeper reason for it ?
Note :- I am considering only 2 Dimensions for simplicity.
 A: I am going to assume that you have not yet studied linear algebra, sorry if it seems as if I am talking down to you at any point.
You are correct in that we can split a vector into two components in the plane.  This is because any two linearly independent(not parallel or anti-parallel) vectors form a basis(a set of vectors from which you can "build" other vectors through addition and scalar multiplication) for $\mathbb{R}^2$ (and in general n vectors for $\mathbb{R}^n$).  One can "do physics" in any one of these coordinate systems, but as you said, the mutually perpendicular(orthogonal) ones are much easier to compute with.  A large part of this is because of "independence".  In Newtonian mechanics we often use the idea that movement in the x and the y (and the z, $x_4,x_5, \dots$)  are independent of the motion in the other ones (i.e. we can apply newtons laws to each force separately).  
Now imagine that we had a non-orthogonal basis set of vectors, for example the x-axis and the vector counter-clockwise at $45^\circ$.  If one takes an arbitrary point, then one cannot move it along one axis without also moving it along the other as well.  So independence of motion is gone, and we would in general end up with a much more complicated system of differential equations than in the orthonormal case.
This idea also comes in to play at the higher level when one studies Fourier series.  This consists of writing a function in terms of complex exponentials.  This is because the complex exponentials form an orthogonal basis for $L^2$, the space of absolutely square integrable functions.  The space $L^2$ is very important, as one of the axioms of quantum mechanics is that all wavefunctions belong to it(and moreover are unit vectors in it).
In fact any Hilbert space(a space where convergent sequences stay in the space and the space has  a "dot product") admits an orthogonal basis.  This is a consequence of Zorn's Lemma(an equivalent statement to the axiom of choice).  Questions such as these belong to functional analysis, linear algebra's infinite dimensional cousin.           
EDIT: This is in response to the comment asking to explain "independence" more.
Lets call our original orthogonal frame of reference $S$ and our new basis $S^\prime$. The coordinates of a point in the difference coordinate system $\xi_S, \xi_{S^\prime}$ these are related by
\begin{align}
\xi_{S^\prime} = A\xi_S \\
A = (\begin{smallmatrix} a&b\\ c&d \end{smallmatrix})
\end{align} 
So if $\xi_S = (x, y), \xi_{S^\prime} = (\alpha, \beta)$ then we would have that 
\begin{align}
ax + by = \alpha \\
cx + dy = \beta
\end{align}
So if we move along the $\alpha$ or $\beta$ axis, then in general both the $x$ and $y$ will change.  In the case above $x  = \alpha$, so $c\alpha + dy = \beta$.  So if we go along the $\beta$-axis, we must also change the $\alpha = x$ and the $y$ coordinate.  So we can't move along $\beta$ without moving along $\alpha$.     
