How did the Transformation of coordinates take place? Well, I was reading about plane waves in Principles of Optics by Born and Wolf. 
Here, $\mathbf r$ is the position vector of a point $P$ and $\mathbf s$ is a unit vector perpendicular to the plane.
Now, they chose a new set of coordinate axes such that $\mathrm{O\zeta}$ in the direction of $\mathbf s\,.$
Thus, $$\mathbf r\cdot \mathbf s~=~ \zeta\,.$$
But then they wrote $$\frac{\partial}{\partial x} = s_x\frac{\partial}{\partial \zeta}, ~~ \frac{\partial}{\partial y} = s_y\frac{\partial}{\partial \zeta},~~\frac{\partial}{\partial z} = s_z\frac{\partial}{\partial \zeta}\,.$$

How did they transform the coordinates? I'm not getting how they wrote the differential operator relations above.
Could anyone shed some light on how to derive the relations from $\mathbf r\cdot\mathbf s~=~\zeta$?
 A: Writing $u$ instead of $\zeta$, we first find an expression for the differential of $u$.
$$u=\vec{r}\cdot\hat{s} = s_x x+ s_y y + s_z z \\
du = \frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy + \frac{\partial u}{\partial z}dz  \\
du = s_x dx + s_y dy + s_z dz $$
Now consider the differential of a function $f(u) = f(\vec{r}\cdot\hat{s})$.  Considering $f$ to be a function of $u$ we can write its differential as
$$df=  \frac{\partial f}{\partial u}du = \frac{\partial f}{\partial u}\left( s_x dx + s_y dy + s_z dz \right) \\
df = \left(s_x\frac{\partial f}{\partial u}\right) dx + \left(s_y\frac{\partial f}{\partial u}\right) dy + \left(s_z\frac{\partial f}{\partial u}\right) dz \;.$$
Now consider $f$ to be a function of $x$, $y$, and $z$ and write its differential as
$$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz \;.$$
Now we compare the last two expressions for $df$ and we keep in mind that $x$, $y$ and $z$ can be varied independently.  When we hold $y$ and $z$ constant and vary $x$ the last two expressions for $df$ give us
$$\Delta f= \left(s_x\frac{\partial f}{\partial u}\right) \Delta x=\frac{\partial f}{\partial x} \Delta x\;.$$
Therefore we must have
$$\frac{\partial f}{\partial x} = s_x\frac{\partial f}{\partial u}\;.$$
Since $f$ an arbitrary (differentiable) function, we conclude that the 
$$\frac{\partial}{\partial x} = s_x\frac{\partial}{\partial u} \;,$$
which is the first operator equation in Born & Wolf.  The remaining equations follow by considering what happens when we hold $x$ and $z$ fixed and vary $y$, etc.
To continue beyond your question, the next steps that Born & Wolf do not show could be to write the second derivatives as 
$$\frac{\partial^2}{\partial x^2} = s_x^2\frac{\partial^2}{\partial u^2} \;, \\ \frac{\partial^2}{\partial y^2} = s_y^2\frac{\partial^2}{\partial u^2} \;, \\ \frac{\partial^2}{\partial z^2} = s_z^2\frac{\partial^2}{\partial u^2} \;,\\$$
which leads to
$$\nabla^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2} +\frac{\partial^2}{\partial z^2} = s_x^2\frac{\partial^2}{\partial u^2} + s_y^2\frac{\partial^2}{\partial u^2} + s_z^2\frac{\partial^2}{\partial u^2} \\ = \left( s_x^2 + s_y^2 + s_z^2 \right) \frac{\partial^2}{\partial u^2} = \frac{\partial^2}{\partial u^2} \;,$$
where the last step follows from the fact that $\hat{s}$ is a unit vector.  Therefore, we have Born & Wolf's next equation
$$\nabla^2 V = \frac{\partial^2 V}{\partial \zeta^2}\;.$$
