OK so I’m doing motion of a particle along a parametric curve we call $R(t)$. We have acceleration in terms of a unit normal vector and unit tangent vector. Acceleration at a point is given by $\vec{a} = \frac{d^2s}{dt^2}\vec{T}+\left(\frac{ds}{dt}\right)^2\cdot\frac{d\phi}{ds}\cdot\vec{N}$ Where $\vec{T}$ and $\vec{N}$ are the unit Tangent and normal vectors. My doubts are as follows :
I can understand $\frac{ds}{dt}$ but not $\frac{d^2s}{dt^2}$. How do I visualize it?
I know $\frac{d\phi}{ds}$ is the rate of change of $\phi$ with respect to arc length. But how do I perceive that?
Finally we have $\vec{T}=\cos\phi\hat{\imath} + \sin\phi\hat{\jmath}$ and we say $\frac{d\vec{T}}{d\phi}$ is $\vec{N}$. But we can write it also as $\vec{T}=\frac{dx}{dt}\hat{\imath}+\frac{dy}{dt}\hat{\jmath}$. Now suppose I want to differentiate this with respect to ϕ then what? Also how do I visualize this because there are so many terms like $\frac{d}{d\phi}(\frac{dx}{dt})$.
$\newcommand{\Vec}[1]{\mathbf{#1}}$To address your first two questions, let's assume your curve is regular, i.e., that the velocity $R'$ is nowhere-vanishing. Writing $$ R'(t) = \|R'(t)\|\, \frac{R'(t)}{\|R'(t)\|} = v(t) \Vec{T}(t) $$ uniquely expresses the velocity vector as a magnitude (the speed $v(t) = \frac{ds}{dt}$) and a direction (the unit vector $\Vec{T}(t)$).
Differentiating again, using the product rule, gives $$ R''(t) = v'(t) \Vec{T}(t) + v(t) \Vec{T}'(t). \tag{*} $$ The first summand is obviously parallel to the velocity. The second is orthogonal to the velocity: differentiating $\Vec{T}(t) \cdot \Vec{T}(t) = 1$ gives $\Vec{T}(t) \cdot \Vec{T}'(t) = 0$.
In words, (*) decomposes the acceleration into tangential and normal components. Respectively, these terms correspond to the change in speed $v'(t) = \frac{d^{2}s}{dt^{2}}$ (along the direction of motion) and the change in direction. (If $\Vec{T}'(t_{0}) \neq 0$, there is a unique circle lying in the plane through $R(t_{0})$ and spanned by $\Vec{T}(t_{0})$ and $\Vec{T}'(t_{0})$ that agrees with $R(t)$ to second order. If this circle has radius $r$, it turns out that $v(t_{0}) \|\Vec{T}'(t_{0})\| = \frac{v(t_{0})^{2}}{r}$, a well-known formula for centripetal acceleration.)
All of the preceding carries through for arbitrary regular curves, not just plane curves.
As for the third item, for a Euclidean plane curve the unit vector $\Vec{T}$ may be written as $(\cos\phi) \Vec{i} + (\sin\phi) \Vec{j}$, in which case $$ \frac{d\Vec{T}}{d\phi} = (-\sin\phi) \Vec{i} + (\cos\phi) \Vec{j} $$ is orthogonal to $\Vec{T}$. (This usage of Leibniz notation is likely to cause grief, as you've already noticed, though it expresses a simple idea: When a unit vector rotates, its "rate of change with respect to angle" is an orthogonal unit vector.)
The formula $\Vec{T}(t) = \frac{dx}{dt}\, \Vec{i} + \frac{dy}{dt}\, \Vec{j}$ is not correct unless the curve happens to have unit speed.