computational insight behind why connections fix the shape of surface Based on a video lecture, I had some queries. If we just have a manifold [M-set,O-topology,A-atlas] say $S^2$, this manifold represents a football or a potato equally. But once we choose a connection $\nabla$ on it, it will only represent say an ellipsoid and nothing else. It's shape is now fixed. No stretching or deformation is now allowed. 
Can you explicitly define a connection on $S^2$ and show why it is fixing the shape of the manifold computationally (Even the example of a spherical connection will do)?
Conversely, given a shape of the manifold, how can its connection be written down explicitly from that piece of information? 
Maybe the latter question makes more sense than the former question!
It seems clear to me in respect of real functions $f:\mathbb{R}\rightarrow \mathbb{R}$ that if $f'$ is given and some initial conditions, the shape of curve $"f(x)\,\,vs\,\,x"$ is fixed. But how does this generalize in terms of manifold and connection? What is the underlying physical idea/intuition behind this relationship of an abstract concept viz. connection and the physical shape? This must have to do with the very early days when connection was defined. But modern expositions of connection don't go into all that and it difficult to find those ideas.
EDIT:
Based on Lee Mosher's comment i shortened the question and will ask other queries later in other questions.
 A: Since you're interested in the last part, I'll skim through the rest and try my best to convey what was done in the final part of that lecture, filling in the gaps with the calculations that Schuller had to omit for obvious time reasons.
Preliminary work. Suppose we have already visualized $\mathbb S^2$ as the set of its points, i.e. the triples of real numbers such that the condition $a^2+b^2+c^2 = 1$ is satisfied:
$$\mathbb S^2 := \{(a,b,c) \in \mathbb R^3\ |\ a^2+b^2+c^2=1 \} $$
and endowed it with the standard topology
$$\mathscr O_{\mathrm{st},\mathbb S^2} = \{\mathcal W \subseteq \mathbb S^2\ |\ \exists \mathcal U \in \mathscr O_{\mathrm{st},\mathbb R^3}\ \mathrm{s.t.}\ \mathcal W = \mathcal U \cap \mathbb S^2 \} $$
As I said above, the atlas $\mathscr A_{\mathrm{pr}}$ we are working with consists of charts
$$\begin{split}
(\mathcal U, x) &: \begin{cases} \mathcal U = \mathbb S^2 \setminus (0,0,-1) \\
x : \mathcal U \to \mathbb R^2, & x(a,b,c) = \left(\dfrac{2a}{1+c},\dfrac{2b}{1+c}\right) \end{cases} \\
(\mathcal V,y) &: \begin{cases} \mathcal V = \mathbb S^2 \setminus (0,0,+1) \\
y : \mathcal V \to \mathbb R^2, & y(a,b,c) = \left(\dfrac{2a}{1-c},\dfrac{2b}{1-c}\right) \end{cases}
\end{split}$$
The map $y$ may be visualized by sticking a round sphere onto $\mathbb R^2$ so that the south pole coincides with the origin; then dropping a ray from the north pole downward so that it intersects both the sphere and the plane. Indeed you can check that, e.g., $y$ maps the south pole to the origin of $\mathbb R^2$ ($y : (0,0,-1) \mapsto (0,0)$), and the equatorial circle to the circle of radius $2$ in the plane. The map $x$ is the same, with north and south poles swapped.
Intuitively, $x$ and $y$ are both invertible maps; inverting them corresponds to "folding" the plane back onto the sphere. Algebraically, one can see that the inverse maps $x^{-1} : x(\mathcal U) \to \mathcal U$ and $y^{-1} : y(\mathcal V) \to \mathcal V$ amount to
$$\begin{split}
x^{-1}(\alpha,\beta) &= \left(\frac{4\alpha}{\alpha^2 + \beta^2 + 4},\frac{4\beta}{\alpha^2+\beta^2+4},-\frac{\alpha^2+\beta^2-4}{\alpha^2+\beta^2 + 4}\right) \\
y^{-1}(\gamma,\delta) &= \left(\frac{4\gamma}{\gamma^2 + \delta^2 + 4},\frac{4\delta}{\gamma^2+\delta^2+4},+\frac{\gamma^2+\delta^2-4}{\gamma^2+\delta^2 + 4}\right)
\end{split} $$
For both maps, we can retrieve the identity of the domain or the range by direct substitution of one formula into its inverse.
It can be seen that $x$ and $y$ are homeomorphisms, but we want to establish if the transition maps
$$\begin{split}
(x \circ y^{-1}) &: x(\mathcal U \cap \mathcal V) \to y(\mathcal U \cap \mathcal V) \\
(y \circ x^{-1}) &: y(\mathcal U \cap \mathcal V) \to x(\mathcal U \cap \mathcal V)
\end{split}$$
are more than continuous (namely infinitely differentiable). Indeed, we see that
$$(x\circ y^{-1}) : (\gamma,\delta)\mapsto\left(\frac{2\left(\frac{4\gamma}{\gamma^2+\delta^2+4}\right)}{1+\left(\frac{\gamma^2+\delta^2-4}{\gamma^2+\delta^2+4}\right)},\frac{2\left(\frac{4\delta}{\gamma^2+\delta^2+4}\right)}{1+\left(\frac{\gamma^2+\delta^2-4}{\gamma^2+\delta^2+4}\right)}\right) = \left(\frac{4\gamma}{\gamma^2+\delta^2},\frac{4\delta}{\gamma^2+\delta^2}\right) $$
and, with similar computations,
$$(y\circ x^{-1}) : (\alpha,\beta) \mapsto \left(\frac{4\alpha}{\alpha^2+\beta^2},\frac{4\beta}{\alpha^2+\beta^2}\right)$$
meaning that, as we expected, the transition maps are $\mathcal C^\infty$, so that $\mathscr A_{\mathrm{pr}}$ is actually a smooth atlas for the sphere. This is geometric proof that we are dealing with the "smooth sphere" $(\mathbb S^2,\mathscr O_{\mathrm{st},\mathbb S^2}, \mathscr A_{\mathrm{pr}})$ and not, for example, with the surface of a cube or a dodecahedron.
Establishing a metric. Define the following four (smooth) maps (technically, smooth scalar fields): for $i,j = 1,2$,
$$g_{(x)ij} : \mathbb S^2 \to \mathbb R \quad g_{(x)ij}(a,b,c) = \begin{cases} \frac{16}{\left(\left(\frac{2a}{1+c}\right)^2+\left(\frac{2b}{1+c}\right)^2+4\right)^2} & i = j \\ 0 & i \neq j\end{cases}$$
and similarly
$$g_{(y)ij} : \mathbb S^2 \to \mathbb R \quad g_{(y)ij}(a,b,c) = \begin{cases} \frac{16}{\left(\left(\frac{2a}{1-c}\right)^2+\left(\frac{2b}{1-c}\right)^2+4\right)^2} & i = j \\ 0 & i \neq j\end{cases}$$
Looks complicated, but this should be defining the coefficients of (a scalar multiple of) the "standard" metric tensor $g \in \mathscr T^0_2 \mathbb S^2$, as seen from the charts $x$ and $y$ respectively.
Indeed we may check that the above-defined functions $g_{(x)ij}$ and $g_{(y)ij}$ are the chart-representations of a tensor $g$, by using the fact that
$$g_{(x)ij} = \frac{\partial y^m}{\partial x^i} \frac{\partial y^n}{\partial x^j} g_{(y)nm}$$
and we may check that they represent the very same metric, by seeing that, for any choice of vector fields $X,Y$, the number $g(X,Y)$ is independent of the choice of charts:
$$g(X,Y) = g_{(x)ij}X^i_{(x)}Y^j_{(x)} = \frac{\partial y^m}{\partial x^i} \frac{\partial y^n}{\partial x^j} g_{(y)nm} \frac{\partial x^i}{\partial y^s} X^s_{(y)} \frac{\partial x^j}{\partial y^t} Y^t_{(y)} = g_{(y)nm} X^s_{(y)} Y^t_{(y)} \delta^m_s\delta^n_t = g_{(y)nm} X^m_{(y)} Y^n_{(y)} $$
So we are sure that a tensor field $g$ exists, such that its chart representations are the above defined functions. This endows all vector structures on $\mathbb S^2$ with the notion of "vector length" (given by the square root of $g(X,X)$, if $X$ is a vector or vector field), which allows us to talk about the "speed" of the parametrization of a curve. For example, let $\gamma : \mathbb R \to \mathbb S^2$ such that $\gamma(\lambda) = (\cos\lambda,\sin\lambda,0)$: we have that the vector tangent to $\gamma$ at the point corresponding to $\lambda = 0$, written $X_{\gamma,\gamma(0)}$, is given in coordinates by
$$X_{\gamma,\gamma(0)} = X^i_{(x)\gamma,\gamma(0)}\frac{\partial}{\partial x^i}\Big|_{\gamma(0)} = (x^i \circ \gamma)'(0)\frac{\partial}{\partial x^i}\Big|_{(1,0,0)} = 2\frac{\partial}{\partial x^2}\Big|_{(1,0,0)}$$
so that we have
$$g(X_{\gamma,\gamma(0)},X_{\gamma,\gamma(0)}) = g_{(x)ij}X^i_{(x)\gamma,\gamma(0)}X^j_{(x)\gamma,\gamma(0)} = g_{(x)22}X^2_{(x)\gamma,\gamma(0)} = \frac{16}{\left(\left(2\right)^2+0+4\right)^2} \cdot 2 = 1 $$
Thus the speed of the parametrization of $\gamma$ is unitary, as we would have expected. The metric also allows us to talk about the length of a curve, which is a hint toward the fact that establishing a metric also implicitly fixes the shape of our manifold. In our case, we may even provide an induced connection that shows this explicitly.
Constructing the connection. The connection is uniquely established, in the coordinates $x$, by the $2^3$-many connection coefficients $\Gamma_{(x)bc}^a$. The round sphere is then given by
$$\Gamma^i_{jk} = \frac 1 2 \tilde g^{im} \left(\frac{\partial}{\partial x^j} g_{mk} + \frac{\partial}{\partial x^k} g_{jm} - \frac{\partial}{\partial x^m} g_{jk}\right)$$
where $\tilde g$ is the unique $\binom{2}{0}$-tensor field (the "inverse metric") satisfying $\tilde g^{mk}g_{m\ell} = \delta^k_\ell$. This particular connection (the Levi-Civita connection) is the one that defines the straightest curves to be the fastest, according to the definition of speed given by the metric. And this is the sense in which the connection fixes the shape of a manifold: on an ellipsoid, or a potato, the fastest route between two points is intuitively different than what it would be on a round sphere, and the choice of a similar metric-based connection on the ellipsoid would establish this alternative route to be the straightest.
(It may be checked that the curve $\gamma$ already defined above is in fact autoparallely transported according to the Levi-Civita connection! This is in line with the fact that it represents the equator of our sphere – a great circle – with uniform parametrization.)
