Consider $S^n\cong SO(n+1)/SO(n)$. Thus we have an homogeneous space, whose isometry group is $SO(n+1)$. We have the round metric (the tensor preserved by the isometry group) and the generators of this group are the $\frac{n(n+1)}{2}$ Killing vector fields $v_{ij}$, antisymmetric in $i,j=1,\ldots,n+1$.
One can embed the sphere in $\mathbb{R}^{n+1}$ by spherical coordinates: $$x_i = R y_i(\alpha_j)$$
The $y_i$ are functions of the angles $\alpha_j$ (they are products of sines and cosines) and they satisfy the relation $y_iy_i = 1$ (sum over repeated indices).
With these notations I can write explicitly the Killing vectors as follows,
$$v_{ij} = y_i k_j - y_j k_i$$
where $k_i$ are the Conformal Killing vectors, i.e.
$$\mathcal{L}_{k_i} g = - y_i g\, .$$
The question now is the following,
given the algebra of $\mathfrak{so}(n+1)$ satisfied by $v_{ij}$, is there any algebra for the Conformal Killing vectors?
The conformal Killing vectors on $S^n$ form the Lie algebra $\mathfrak{so}(n+1,1)$ of a Lorentzian orthogonal group. The best way to see this is to identify $S^n$ with the projectivized Light cone (i.e. the space of isotropic lines) in the Lorentzian vector space $\mathbb R^{(n+1,1)}$. If you realize this as points $(x,t)$ with $\langle (x,t),(y,s)\rangle=\sum_ix_iy_i-ts$, you see that $(x,t)$ is isotropic if and only if the Euclidean norm of $x$ equals $|t|$. Hence any isotropic line contains a unique point $(x,1)$ with $x\in S^n$.
Then the standard action of orthogonal group $O(n+1,1)$ on $\mathbb R^{(n+1,1)}$ maps isotropic lines to isotropic lines and thus descends to an action on $S^n$ which is easily seen to be by conformal isometries. Infinitesimally you get a Lie algebra of conformal Killing fields isomrophic to $\mathfrak{so}(n+1,1)$.