Suppose we have a Riemann sphere expressed as $S^2:=\{ (x,y,z)\in \mathbb{R}^3: x^2+y^2+z^2=1\}$ equipped with a standard inner product in $\mathbb{R}^3$ since $$ T_{(x_0,y_0,z_0)}S^2 = \{ (x,y,z) \in \mathbb{R}^3: xx_0+yy_0+zz_0=0 \} $$
In short, I want all differential calculations on $S^2$ to become the ones in $\mathbb{R}^3$. Suppose $\nabla$ is the corresponding Riemann connection, is there a general formula for $$ \nabla_{(x_1,y_1,z_1)}(x_2,y_2,z_2) $$
in a selected $T_{(x_0,y_0,z_0)}S^2$?
This is explained in a textbook or course on the differential geometry of curves and surfaces (e.g., Do Carmo's book).
First, you cannot take the covariant derivative of a single tangent vector $(x_2,y_2,z_2)$. You need a vector field. Second, a tangent vector field on $S$ is a map $V: S \rightarrow \mathbb{R}^3$ that happens to satisfy $V(x) \in T_xS$. Given a point $(x_0,y_0,z_0) \in S$, the covariant derivative of $V$ in the direction $w =(x_1,y_1,z_1)$ is the orthogonal projection of the directional derivative $D_wV(x_0,y_0,z_0)$ onto $T_{(x_0,y_0,z_0)}S$.
If $S$ is the unit sphere centered at the origin, then $$T_{(x_0,y_0,z_0)}S = \{ v \in \mathbb{R}^3\ :\ v\cdot x = 0 \}.$$ It is now straightforward to find an explicit formula for the covariant derivative of a tangent vector field on $S$.