A Universe of Sorts

§ Connections, take 2

I asked a math.se question about position, velocity, acceleration that recieved a great answer by peek-a-boo. Let me try and provide an exposition of his answer.
Imagein a base manifold $M$ , say a circle.
Now imagine a vector bundle over this, say 2D spaces lying above each point on the circle. Call this $(E, \pi, M)$
What is a connection? Roughly speaking, it seems to be a device to convert elements of $TM$ into elements of $TE$ .
We imagine the base manifold (circle) as horizontal, and the bundle $E$ as vertical. We imagine $TM$ as vectors lying horizontal on the circle, and we imagine $TE$ as vectors lying horizontal above the bundle. So something like:

So the connection has type $C: E \times TM \to TE$ . Consider a point $m \in M$ in the base manifold.
Now think of the fiber $E_m \subseteq E$ over $x$ .
Now think of any point $e \in E_m$ in the fiber of $m$ .
This gives us a map $C_e: T_e M \to T_e E$ , which tells us to imagine a particle $e \in E$ following its brother in $m \in M$ . If we know the velocity $\dot m \in T_m M$ , we can find the velocity of the sibling upstrairs with $C_e(\dot m)$ .
In some sense, this is really like path lifting, except we're performing "velocity lifting". Given a point in the base manifold and a point somewhere upstairs in the cover (fiber), we are told how to "develop" the path upstairs given information about how to "develop" the path downstairs.
I use "develop" to mean "knowing derivatives".

§ Differentiating vector fields along a curve

Given all of this, suppose we have a curve $c: I to M$ and a vector field over the curve $v: I \to E$ such that the vector field lies correctly over the curve; $\pi \circ v = c$ . We want to differentiate $v$ , such that we get another $v': TI \to E$ .
That's the crucial bit, $v$ and $v'$ have the same type, and this is achieved through the connection. So a vector field and its derivative are both vector fields over the curve.
How do we do this? We have the tangent mapping $Tv: TI \mapsto TE$ .
We kill off the component given by pushing forward the tangent vector $Tc(i): TI$ at the bundle location $v(i)$ via the connection. This kills of the effect of the curving of the curve when measuring the change in the vector field $v$ .
We build $[z(t_i: TI) \equiv Tv(ti) - C_{v(i)}(Tc(i))]: TI \to TE$ .
We now have a map from $I$ to $TE$ , but we want a map to $E$ . What do?
Well, we can check that the vector field we have created is a vertical vector field, which means that it lies entirely within the fiber. Said differently, we check that it pushes forward to the zero vector under projection, so $TM: TE \to TM$ will be zero for the image of $w$ .
This means that $z$ lies entirely "inside" each fiber, or it lies entirely in the tangent to the vector space $\pi^{-1}(m)$ (ie, it lives in $T\pi^{-1}(m)$ ), instead of living in the full tangent bundle $E_m$ where it has access to the horizontal components.
But for a vector space, the tangent space is canonically isomorphic to the vector space itself! (parallelogram law/can move vectors around/...). Thus, we can bring down the image of $w$ from $TE$ down to $E$ !
This means we now have a map $z: TI \to E$ .
But we want a $w: I \to E$ . See that the place where we needed a $TI$ was to produce