Basic question: intuition about $W = F \cdot T$ I find it written many places that "you can find the work along a short segment of the path by taking the dot product of the force and the tangent vector."
I can solve these problems, but I have a question about the intuition. Why do I use the unit tangent vector? Is it because the distance being traveled is so infinitesimal that the tangent vector (a straight line, informally put) can approximate the curve? I suppose this makes the calculation easier? And why do we use the "unit" tangent vector? Again, I know how to find the unit tangent, and I understand what it represents generally. But why do we take the dot product of the force and the unit tangent vector, as opposed just to the 'tangent vector'?
So my two questions are 1) why use the tangent vector, as opposed to some other way to estimate arc length; and 2) why the unit tangent vector?
I apologize in advance for the simple (simplistic) nature of these questions. I'm fine to solve these problems, but I really want to understand the "why".
 A: The work over a short segment of path is the dot product between force $\vec F$ and small displacement $\delta r$.
$$\delta W=\vec F \cdot \delta\vec r$$
Now, the work over the entire path (i.e. not very small displacement) will be the sum of the work done over those infinitesimally small displacements. So what you end up doing is calculating the following integral $$W_{\rm by\; \vec{F}} \equiv \int_C \vec F \cdot d\vec r \tag1$$
Now recall that velocity is always tangent to the path. Since $\vec v \equiv d\vec r/dt$, then you have that $\delta \vec r = \vec v\; \delta t$. Since $\delta t$ is a scalar, this implies that $\delta \vec r$ and $\vec v$ are parallel. So $d\vec r$ is tangent to the path.

why use the tangent vector, as opposed to some other way to estimate arc length

Taking the $\vec F \cdot d\vec r$ dot product gives us the projection of $\vec F$ along the displacement, multiplied by the length of that small displacement. In other words, this dot product gives us the component of $\vec F$ that is acting either parallel or antiparallel to the displacement (multiplied by the length of that small displacement).
Therefore, work is also given by $W_{\rm by\; \vec{F}} \equiv \int_C F_\parallel \; ds$ where $ds=|d\vec r|$. This is because when calculating work, we only care about the force that acts in the direction of/against the motion. This is due to the fact that work is done to try to speed up/slow down an object. Changing the object's path without changing the object's speed will produce no net work.

why the unit tangent vector

Work is not simply the tangent vector dotted with the force. The length of the small displacement actually matters (because if you move an object of mass $m$ over a distance $s$ and then that same object is moved over that same distance but this time it's moved faster, then more work is done).
With that said, it should be clear that work is not simply $\vec F \cdot \vec T$ -- we must take into account how the object sped up over some displacement.
You may be confusing that $F\cdot T$ with the following (correct) equation:
$$W_{\rm by\; \vec{F}} \equiv \int_C \vec F \cdot \vec T \;ds\tag2$$
Recall that $ds=|d\vec r|$ and that the unit tangent is $\vec T=\dfrac{\vec v}{|\vec v|}=\dfrac{d\vec r}{|d\vec r|}\implies \vec T\; ds=d\vec r$. Substituting that into Eq. (1) leads us to Eq. (2).
A: I think the intuition you're looking for is that the contribution of the force to the work is its component in the direction of the path. So if you have force defined as a vector field $ \mathbf F$ and a smooth curve $ s $ you calculate the work over the segment $ [a,b] $ as:
$$ W_{ab} = \int_a^b \mathbf F \cdot  d\mathbf s $$
Note the dot product, this is how we get the component of the force that's in the direction of (tangential to) the curve.
A: Work indeed does have $ \vec{F} \cdot \vec{T}$ but I think you forgot a term as well. The correct expression for infinitesimal work is given as:
$$ dW = \vec{F} \cdot \vec{T} ds$$
However, I think it's easier to evaluate the following equivalent quantity:
$$ dW = \vec{F} \cdot \vec{\frac{dr}{dt} } dt$$
A: So the work done by a force given by the vector field $\vec F$ is
$$W = \vec F \cdot \vec v \ dt$$
If the force varies from point to point, the displacement vector $\vec v dt$ may also change, as the object may follow a curved path in two or three dimensions.
Suppose that the path of an object is given by a vector function $r(t)$ at any point along the path, then the tangent vector $\vec{\dot r}\Delta t$ gives an approximation to its motion over a short time $\Delta t$, so that the work done during that time is approximately $\vec F . \vec{\dot r} \Delta t$
This means then that the total work over some time period is $t$
$$W = \int_{0}^{t} \vec F \cdot \vec{\dot r} \ dt$$
$\vec{\dot r} \ dt$ is a tangent vector since it touches the curve described by $\vec r (t)$ at a point where we are calculating the instantaneous work done at that point in the infinitesimal interval $\vec{\dot r} dt$.
Note that there is no need here for the use of a unit vector in place of the tangent vector since work is not the dot product of the force and unit vector. Rather, it is the dot product of the force $\vec F$ and the distance interval given by
$$\vec{\dot r} dt = \frac{ \vec {dx}}{dt} dt =  \vec {ds}$$
so that work done
$$W = \int \vec F \cdot \vec{ds}$$
in general. The dot product above ensures that we calculate the product of the component of the force (in the direction of) parallel to the direction of the displacement.
