From my understanding, the tangent space at the identity of the homogeneous space $\rm SO(3) / \rm SO(2)$ is just the quotient space $\mathfrak{so}(3) / \mathfrak{so}(2)$. An element in this quotient space is an element of one of the left cosets {$gh \mid g \in \mathfrak{so}(3), h \in \iota(\mathfrak{so}(2))$} where $\iota : \mathfrak{so}(2) \mapsto \mathfrak{so}(3)$ is a natural embedding. An arbitrary element I'm guessing can be written as $$ \mathfrak{so}(3) / \mathfrak{so}(2) \ni E = \begin{bmatrix} 0 & -c & b \\ c & 0 & -a \\ -b & a & 0 \end{bmatrix} \begin{bmatrix} 0 & -d & 0 \\ d & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -cd & 0 & b \\ 0 & -cd & -a \\ ad & bd & 0 \end{bmatrix} $$
I also know that $\text{SO}(n)$ acts on $\mathbb{R}^n$ preserving the standard inner product, and this naturally induces a transitive action on the unit sphere $S^{n-1}$ with stabilizer $\text{SO}(n-1)$, meaning there is an isomorphism $$S^{n-1} \cong \text{SO}(n) / \text{SO} (n-1)$$ Hence, $\rm SO(3) / \rm SO(2)$ can be interpreted as $S^2$, and $\mathfrak{so}(3) / \mathfrak{so}(2)$ representing an arbitrary tangent space on $S^2$.
My question is, (assuming we choose a random point on $S^2$ as the "identity"), if we embed $S^2$ into $\mathbb R^3$, and say choose the point $(R, 0, 0)$ as the identity, is there a way to convert the matrix $$\begin{bmatrix} -cd & 0 & b \\ 0 & -cd & -a \\ ad & bd & 0 \end{bmatrix} $$ into an actual "geometric" tangent vector at $(R, 0, 0)$ of the sphere in the form $\langle 0, y, z\rangle$?
Consider $\Bbb S^n\subseteq \Bbb R^{n+1}$, write $\Bbb R^{n+1} = \Bbb R^n \times \Bbb R$, and fix the north pole $p = (0,1)$. Note that $T_p(\Bbb S^n) = \Bbb R^n\times \{0\}$. Then ${\rm SO}(n+1)$ acts on $\Bbb S^n$ by evaluation, and the stabilizer of $p$ under this action is the image of the embedding $\iota\colon{\rm SO}(n) \hookrightarrow {\rm SO}(n+1)$ given by $\iota(R)(v,t) = (Rv,t)$. We identify ${\rm SO}(n)$ and $\iota[{\rm SO}(n)]$.
The orbit map ${\rm SO}(n+1) \ni A \mapsto Ap \in \Bbb S^n$ is a smooth submersion, inducing a diffeomorphism ${\rm SO}(n+1)/{\rm SO}(n)\to \Bbb S^n$, whose derivative is an isomorphism $\mathfrak{so}(n+1)/\mathfrak{so}(n) \to T_p(\Bbb S^n)$.
Hence, $T_p(\Bbb S^n)$ is isomorphic to any subspace $\mathfrak{m}$ of $\mathfrak{so}(n+1)$ which is complementary to $\mathfrak{so}(n)$. Here's the most natural $\mathfrak{m}$ we'll ever find: $$\mathfrak{m} = \left\{ \begin{bmatrix} 0_{n\times n} & v \\ -v^\top & 0 \end{bmatrix}\mid v \in \Bbb R^n\right\},$$ Here, $v\in \Bbb R^n$ is treated as a column vector, and $[\,\cdot\,]^\top$ denotes matrix transposition. It is clear that $\mathfrak{so}(n+1) = \mathfrak{so}(n)\oplus\mathfrak{m}$. The mapping $$\mathfrak{m} \ni \begin{bmatrix} 0_{n\times n} & v \\ -v^\top & 0 \end{bmatrix} \mapsto (v,0) \in p^\perp = T_p(\Bbb S^n) \subseteq \Bbb R^n\times \Bbb R $$ is the isomorphism you want.
What makes the situation here extra-special is that $\mathfrak{m}$ is invariant under the adjoint action of ${\rm SO}(n)$, making $\Bbb S^n$ a reductive homogeneous space (it is not hard to check and I think it will be instructive for you to do it).
In general, a reductive decomposition for a smooth homogeneous space $G/H$ is defined as a vector space decomposition $\mathfrak{g} = \mathfrak{h}\oplus \mathfrak{m}$, with the subspace $\mathfrak{m}$ being ${\rm Ad}(H)$-invariant; then $G/H$ is called reductive if it admits a reductive decomposition.
The geometry of a connected Lie group $G$ can be studied without loss of information on the infinitesimal level $\mathfrak{g}$. For a reductive homogeneous space, the same can be done with $\mathfrak{m}$ playing the role of $\mathfrak{g}$. The Lie algebra $(\mathfrak{g},[\cdot,\cdot])$ is replaced with a nonassociative algebra $(\mathfrak{m},[\cdot,\cdot]_{\mathfrak{m}})$, where $[\cdot,\cdot]_{\mathfrak{m}} = {\rm pr}_{\mathfrak{m}}\circ [\cdot,\cdot]|_{\mathfrak{m}\times\mathfrak{m}}$. In addtion, $G$-invariant objects on $G/H$ (tensor fields, distributions, connections) correspond to ${\rm Ad}(H)$-invariant objects on $\mathfrak{m}$ (tensors, subspaces, multiplications). Formulas for curvatures of $G$-invariant metrics on $G/H$ can be computed on $\mathfrak{m}$, etc.
One of my favorite references for this is An Introduction to Lie Groups and the Geometry of Homogeneous Spaces, by A. Arvanitoyeorgos.