What’s an intuitive conceptual way to interpret the formula for a vector perpendicular to a curve? Given a parametrized vector curve in a 2D plane $\newcommand{\r}{\mathbf{R}} \newcommand{\t}{\mathbf{T}} \newcommand{\n}{\mathbf{N}}$
$$\r =
\begin{bmatrix}
x(t) \\
y(t) \\
\end{bmatrix}$$
the tangent vector is
$$\t = \frac{d\r}{dt}$$
This makes total sense to me, because it’s very reminiscent of the tangent line formulations from introductory calculus.
The vector normal or perpendicular to $\r$ is
$$\n = \frac{d\t}{dt} = \frac{d^2\r}{dt^2}$$
This is less obvious to me, and I have trouble visualizing it. It does, however, bring to mind centripetal acceleration being perpendicular to the velocity vector which is tangent to the path of motion. However, I imagine that the math precipitates the physics.
Is there some key concept I’m missing that would make this more intuitive or more obvious?
 A: The second derivative tells you the rate of change of the first derivative. It shows you how fast the curve changes its direction (not how fast it changes its position, which the first derivative tells you) at t, so basically, $N=\frac{\frac{d^2R}{dt^2}}{\big|\frac{d^2R}{dt^2}\big|}$ tells you in which direction the curve "wants to go" at this point.
A: First off, you need to replace “THE vector perpendicular” with “A vector perpendicular” as any scalar multiple of it will also be orthogonal to the curve. Here is the easiest way to see this, 
Let's normalize $\newcommand{\T}{\mathbf{T}} \T$ and rewrite it as $\newcommand{\r}{\mathbf{r}} \T(t) = \r'(t)/ \Vert \r'(t)\Vert$. This vector has the same direction as your version of $\T$ but has unit length. Clearly one will be orthogonal to the curve if and only if the other is.
Note the following equality (the dot here refers to the dot product) :
$$ 1 = \Vert \T(t)\Vert^2 = \T(t) \cdot \T(t) = (\T \cdot \T)(t) $$
We now use the differentiation formula for the dot product and differentiate both sides of the equation:
$$0 = (\T\cdot \T)'(t) = \T(t) \cdot \T'(t) + \T(t) \cdot \T'(t) = 2 \left(\T(t)\cdot \T'(t)\right)\implies \T \cdot \T' = 0$$
Recall two vectors are orthogonal if and only if their dot product is zero. Thus we have shown that $\T'$ is orthogonal to $\T(t)$. If the curve has a degree of curvature (i.e. not a straight line) then $\T' \neq 0$.
Note that $\T'(t) = \r''(t)/\Vert \r'(t) \Vert$ and hence is parallel to $\r''(t)$. 
In conclusion, $\r''(t)$ is orthogonal to the tangent vector and is therefore normal to the curve.

Also, since you expressed interest in the formula for centripetal acceleration, you can simply derive it as follows:
Differentiate the relation $s = r\theta$ to get $v=r\omega$.
Parameterize the path of the object by $$C:\ \r(t):=\bigl\langle r\cos(wt),r\sin(wt)\bigr\rangle\quad t \in [0,1] \quad r = \Vert \mathbf{r} \Vert$$
Differentiate twice to get $$\mathbf{a}(t) = -\omega^2\, \r(t)$$
Thus at every point, the acceleration points in the direction of $-\r$ (i.e. towards the origin/center of the circle).
Using the equation $v=r\omega$ above we get
$$\Vert\mathbf{a}(t)\Vert = \left\|-\omega^2\right\| \, \Vert \r(t) \Vert = \omega^2r\ \left\Vert\bigl\langle\cos(wt),\sin(wt)\bigr\rangle\right\Vert = \omega^2r\times 1 = \omega^2r = \frac{v^2}{r}$$
A: Without evoking physical concepts:
The direction of the second derivative of a planar curve has to do with the so cold osculating circle.
Take a look at the following figure.

Here the red vector is the first derivative of the parametric equation belonging to the black curve and the blue line is the normal line to the same. The normal line contains the second derivative vector which is perpendicular to the red vector. The normal line contains the center of the osculating or "kissing circle" being the best fit to the black curve among the possible tangent circles.
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