Curvature of a space curve: how do we get from the conceptual definition to K = |dT/ds|? (Context, calculus 3)
The definition of curvature I get: "How fast a curve is changing direction at a point" (Source)
But the precise definition(s) I don't understand.  My book has a particularly awful definition and does not define what a "unit tangent vector" is and simply assumes you can figure it out; Forgive me if I'm missing some supporting material.
The definitions I've seen:
$$K = \| \frac{dT}{ds} \|$$
$$K =  \frac{ \| T'(t) \| }{ \| r'(t) \|} $$
I'll take a stab at illustrating my confusion with each.
First: Seems to be the "rate of the change of the tangent unit vector with respect to the arc length".  I suppose this one makes a sort of sense, since you can get an "average" tangent vector by making a secant line between [a,b] over some arc length s and then taking the limit to get an "instantaneous curvature".  But search me why the magnitude of this would be a measurement of curvature.
Second: Seems to be "the magnitude of the derivative of the tangent vector function (r(s(t)) divided by the derivative of the of the vector equation (r(t))" for some reason.  This one I have no idea on.
Wikipedia has a nice writeup and uses circles to define it (which my book does not), it's definition here is particularly helpful:

Given any curve C and a point P on it, there is a unique circle or line which most closely approximates the curve near P, the osculating circle at P. The curvature of C at P is then defined to be the curvature of that circle or line. The radius of curvature is defined as the reciprocal of the curvature.

But then fails to define how this definition leads into the above definitions.
There's one similar post here on stackexchange, but the answer is very confusing:

I won't attempt further diagrams, but now think in 2-dimensions - you will need two different sized circles at every point in order to describe the curvature of the surface at that point.

How and why would you possible need "two different sized circles" to describe curvature?  The figure they use is nearly identical to the wikipedia article and only uses one circle per point, but also fails to describe how we get from there to the definitions above.
For the answer I'd like to request: diagrams using the circle approach, as seen above.  Thank you.
Edit:
Thanks very much to help from whacka in explaining his answer, chatroom log here.  I chose the answer I got from UC Davis as it explains the solution clearly and concisely with a minimum of extra experience required to understand.  I did however give whacka's answer an upvote, since it appears to be popular; I assume because it's a good answer for those with more experience than I have, I'm not fit to judge.
 A: (1) If you're driving and you turn the wheel, the direction that you're facing will change. Indeed, the harder you turn, the faster the direction that you're facing changes. Equally true, your resulting trajectory will be bent in a tighter curve. Thus the rate of change of direction-facing measures curvature of your path.

(2) The curvature of a curve should be independent of one's parametrization of it. The most natural way to parametrize a curve is with the natural parametrization - that is why $\kappa:=\|{\bf T}_s\|$. But how do you compute curvature if you're working with a different parametrization? Why, with the chain rule of course:
$$\frac{{\rm d}{\bf T}}{{\rm d}s}=\frac{{\rm d}{\bf T}/{\rm d}t}{{\rm d}s/{\rm d}t} $$
$\large($Technically the chain rule is $\displaystyle\frac{{\rm d}{\bf T}}{{\rm d}t}=\frac{{\rm d}{\bf T}}{{\rm d}s}\frac{{\rm d}s}{{\rm d}t}$, but just divide that by $\displaystyle\frac{{\rm d}s}{{\rm d}t}$ to isolate the $\displaystyle\frac{{\rm d}{\bf T}}{{\rm d}s}$ we want.$\large)$
Note that $\displaystyle\frac{{\rm d}s}{{\rm d}t}=\left\|\frac{{\rm d}{\bf r}}{{\rm d}t}\right\|$.

(3) At every point on a curve there is a unique tangent line. This line is the best local approximation of the curve to the first order. But a line can be thought of as a generalized circle, one of infinite radius and whose center exists "at infinity," so perhaps consider all circles that are tangent to the curve at a point - which is the best fit? Which approximates it best to the second order?
Without loss of generality, we can apply an affine transformation (translation and rotation of the plane) to move the point on the curve to the origin so that the tangent vector there is on the positive $x$-axis; this greatly simplifies the upcoming algebra. The tangent line, i.e. the $x$-axis itself, interpreted as the graph of a (constant) function, already shares the same first derivative as the curve at the origin. We need to find a circular arc (we only consider the arc of a whole circle so that technically we are talking about a function, at least near the origin) that shares the same second derivative as the curve.
Without loss of generality, suppose the curve, interpreted locally as a graph of $y=f(x)$ near $(0,0)$, has a positive second derivative $f''(0)>0$. Now place a circle on the plane of radius $R$ tangent to the $x$-axis at the origin whose center is above the $x$-axis. The center must be at $(0,R)$. Any small enough arc of this circle through the origin is the graph of the function $g(x)=R-\sqrt{R^2-x^2}$. Compute $g''(0)=R^{-1}$. In order for $f$ and $g$ to have the same second derivative at $0$, we must have $f''(0)=R^{-1}$.
Lo and behold... Parametrize ${\bf r}(t)=(t,f(t))$ so that $\left\|\frac{{\rm d}{\bf r}}{{\rm d}t}\right\|=\sqrt{1+f'(t)^2}$. A tangent vector is given by $(1,f'(t))$, which we can normalize to ${\bf T}(t)=(1+f'(t)^2)^{-1/2}(1,f'(t))$ . Compute
$$\frac{{\rm d}{\bf T}}{{\rm d}t}=-\frac{1}{2}(1+f'(t)^2)^{-3/2}(2f'(t)f''(t))~(1,f'(t))+(1+f'(t)^2)^{-1/2}~(0,f''(t))$$
by the product rule (which works on a scalar function times a vector function) and therefore we have $\|{\bf T}_t(0)\|=f''(0)$, but since $\|{\bf r}_t(0)\|=1$ also we obtain $\|{\bf T}_s\|=\|{\bf T}_t\|/\|{\bf r}_t\|=f''(0)$ at the origin.
Thus we have a verification that $\|{\bf T}_s\|=R^{-1}$.

(4) The question and answer you're linking to is about the curvature of a surface, not of a curve. Granted, the answer begins by explaining the curvature of plane curves, but quickly segues into talking about the Gaussian curvature of surfaces that exist in three-dimensional space. You need to get acquainted with the curvature of a plane curve before you venture into visualizing the curvature of a surface. In fact the quote you give has a typo: it should say you need to think in three dimensions.

Here's another perspective. Consider a circle of radius $R$. First off, if you're driving in a circle at a constant speed, the radius is inversely proportional to how tightly you're turning. The prototypical function of $R$ which is inversely proportional to $R$ is the reciprocal function $R^{-1}$, so it makes sense to make this a definition of the circle's curvature $\kappa$. 
Now say we wanted to compute $R^{-1}$ using only local data of the circle near a point. For instance, if we're in a vehicle and we can measure small changes in speed and direction but this is the only information we have at our disposal. Without loss of generality suppose the center of the circle is the origin, so that it has natural parametrization ${\bf r}(s)=(R\cos\frac{s}{R},R\sin\frac{s}{R})$. Then ${\bf T}(s)=(-\sin\frac{s}{R},\cos\frac{s}{R})$ and $\|{\bf T}_s\|=R^{-1}$.
Heuristically, then, in order to find $R^{-1}$ where $R$ is the radius of the circle that "best-fits" a curve at a point, simply pretend the curve is a circle at that point and compute $\|{\bf T}_s\|$!
A: After some digging on google, I found this from UC Davis, I'll copy the relevant details here for posterity with my own notes.  Feel free to make corrections as my explanations are no doubt not sufficiently precise or may be incorrect in some places, I blame my own ignorance:

Consider a car driving along a curvy road. The tighter the curve, the more difficult the driving is. In math we have a number, the curvature, that describes this "tightness". If the curvature is zero then the curve looks like a line near this point. While if the curvature is a large number, then the curve has a sharp bend.
Before learning what curvature of a curve is and how to find the value of that curvature, we must first learn about unit tangent vector. As the name suggests, unit tangent vectors are unit vectors (vectors with length of 1) that are tangent to the curve at certain points. Because tangent lines at certain point of a curve are defined as lines that barely touch the curve at the given point, we can deduce that tangent lines or vectors have slopes equivalent to the instantaneous slope of a curve at the given point. In other words:

$$ \vec{T}^{\,} = \frac{dr}{dt} $$
I made a small change from the source material here here to illustrate that T is a vector.  r(t) (the parametric equation of some curve) should get you a vector, so r'(t) should also return a vector. dr/dt is the derivative of r(t), in this case the "velocity" vector of the curve at the point t; I hadn't been exposed to thinking about derivatives of r(t) as velocity and acceleration  (etc) prior to this.

which  means

$$ \hat{T} = \frac{\vec{T}^{\,}}{\| \vec{T}^{\,} \|} = \frac{dr/dt}{\| dr/dt \|}  $$
Where we make a unit vector ($\hat{T}$) by dividing $\vec{T}^{\,}$ by its magnitude.  The dr/dt notation is just "unpacking" the definition of $\vec{T}^{\,}$.

Based on what we learned previously, we know that $dr/dt=v$, where v is the velocity at which a point is moving at a given time. Furthermore, the absolute value of the velocity vector is the speed vector of the curve, meaning $|dr/dt|=ds/dt$.

Here the page's notation becomes difficult to follow, simply put $|dr/dt|$ is the absolute value of $dr/dt$, while $\| dr/dt \|$ is the magnitude.  I've made an effort to address that here, but if you were to follow the link (above) you would see no difference in the notation for magnitude and absolute value.

So the formula for unit tangent vector can be simplified to:

$$ \hat{T} = \frac{velocity}{speed} = \frac{dr/dt}{ds/dt}$$
I can't make heads or tails of what follows (some sort of discussion about the "unit tangent vector in terms of arc length"), but we'll be using a definition in the next part, so in summary:
$$ \hat{T} = \frac{dr}{ds} $$
Which we can arrive at from our above equation by "canceling" the dt's:
$$ \hat{T} =\frac{dr/dt}{ds/dt} $$
simplify the complex fraction:
$$ \hat{T} =\frac{dr}{dt} * \frac{dt}{ds} = \frac{dr}{ds} $$

With this information, we will be learning what curvature really is and how we can calculate the curvature, denoted as κ (kappa).
Curvature is a measure of how much the curve deviates from a straight line. In other words, the curvature of a curve at a point is a measure of how much the change in a curve at a point is changing, meaning the curvature is the magnitude of the second derivative of the curve at given point (let's assume that the curve is defined in terms of the arc length s to make things easier).

$$ \kappa = \| \frac{d^2r}{ds^2} \| $$
The wording is a little weird, but I believe I can sum it up:

*

*We get the direction that the curve is curving from $\hat{T}$ (the unit tangent vector)


*If we then find out how fast the direction of $\hat{T}$ is changing at a given point, we know the curvature.
Since if the direction is changing rapidly (a tight curve) our $\kappa$ is large and if our direction is hardly changing at all (a straight line or shallow curve) then our $\kappa$ is small.  Which describes what behavior we want from  $\kappa$.
Essentially the magnitude of the "acceleration" of the curve at a given point is its curvature.
What follows in the discussion is some tedious algebra to make the equations "nicer" to compute with, summarized here:
$$ \kappa = \frac{1}{\| v \|} \| \frac{d\hat{T}}{dt} \| $$
(So we can get $\hat{T}$ in terms of t)
$$ \kappa(t) = \frac{\| r'(t) \times r''(t)\|}{\| r'(t) \|^3} $$
(then stricly in terms of r(t))
$$ \kappa(t) = \frac{\| \hat{T} '(t) \|}{\| r'(t) \|} $$
(then in terms of T and r(t).  The Latex is a little wonky for the notation of "derivative of T-hat" [numerator].)
