We know projection of u onto v is the vector
$proj_{v}u=(|u|cos\theta).\dfrac{v}{|v|}$
The directional derivative f(x,y) is differerntiable in an open region is $ \nabla f .u =|\nabla f||u|cos \theta $
$\because $u is unit vector $|u|=1$
$=|\nabla f|cos \theta$
If , yes then how it linked to rate change of f with arc length in particular .Any intutive way to understand .
You might be mixing up apples (scalars) and oranges (vectors): we project a vector onto a vector and get a vector, but the directional derivative is a scalar. Still, it's a good question because the formulas for scalar projection and the dot product version of the directional derivative do look similar. This isn't coincidental.
The scalar projection of $\mathbf{v}$ onto the nonzero vector $\mathbf{u}$ is "defined" to be $$ \mathrm{sproj}_{\mathbf{u}}\mathbf{v}= |\mathbf{v}| \cos\theta $$ where $\theta$ is the angle between $\mathbf{v}$ and $\mathbf{u}$. (And if $\mathbf{v}$ is the zero vector it hardly matters what you make $\theta$.) This is also the same thing as $$ \mathbf{v} \cdot \frac{\mathbf{u}}{|\mathbf{u}|}. $$ Either way, it's the length of the vector projection when acute, and its negative when obtuse. And, this is all simpler when $\mathbf{u}$ is a unit vector.
If you believe all this, then the directional derivative is the scalar projection of the gradient onto the unit vector $\mathbf{u}$. This comes from the identity $$ D_\mathbf{u}f = \nabla f \cdot \mathbf{u} = |\nabla f| \cos \theta = \mathrm{sproj}_{\mathbf{u}}\nabla f $$ where $\theta$ is the angle between $\nabla f$ and $\mathbf{u}$.