How do you reconstruct the surface knowing the connection coefficients?

In this lec-7 on Connections by Frederic Schuller, the connection is introduced as an extension of the directional derivative of scalar fields to $(p,q)$ tensor field. The initial axioms the connection obeys is same as all that which the directional derivative for scalar field obeys. About 35:07, he says that the remaining freedom of how the directional derivative generalizes is determined by the Christoffel symbols

At 6:30 of lec-8, it is said that the choice of christoffel symbols , fixes our how our manifold looks in space $(M,O , A, \nabla)$. The point I am confused is at 46:26, he concludes the Christoffels of the regular round sphere are given as:

$$ \Gamma_{22}^1= - \sin \theta \cos \theta$$

$$ \Gamma_{12}^2= \Gamma_{21}^2 = \cot \theta$$

With all other $\Gamma$ s vanishing.

My question is how did he figure out these are the correct Christoffel's corresponding to the shape of manifold?

So far in the course, there was no metric introduced, so you can't use the metric tensor formula to find the symbols.

Notes for the lectures can be found here

There are no "correct Christoffel symbols" without a metric. In section 8 of the notes you linked, there is also elaboration of this example. There he showed that upon choosing

$$ \Gamma_{22}^1= - \sin \theta \cos \theta$$ $$ \Gamma_{12}^2= \Gamma_{21}^2 = \cot \theta$$

any great circle satisfies the geodesic (he calls this autoparallel) equation. We can do the converse where the Christoffel symbols can be derived from the geodesic equation once we assume that geodesics on $S^2$ are the great circles. Of course the geodesics themselves will restrict the metrics you can put on $S^2$.