Firstly the bosonic part of superconformal algebra of this theory is $SO(4,2)\times U(1)_R$ not $SO(4,1)\times U(1)_R$. I have corrected this mistake that I guess it was a typo. Now, let us consider the conformal algebra generators.

- The momenta $P^{\mu}$ that generate spacetime translations.
- The angular momenta $J^{\mu \nu}$ that generate spacetime rotations.
- The dilatation operator $D$ that generates uniform rescalings to coordinates and fields.
- The special conformal transformations $K^{\mu}$.

There are many places where you can check the algebra they create, namely the superconformal algebra $\mathfrak{so}(4,2)$. How can this be true? You can verify this by introducing indices $a,b=-1,0,1,2,3,4$ and defining the operators $J^{ab}= -J^{ba}$ as following $J^{ab}= J^{\mu \nu}$ for $a,b=0,1,2,3$, $J^{4\mu}=-J^{\mu4} = \frac{P^{\mu} + K^{\mu} }{2}$, $J^{-1\mu}=-J^{\mu-1} = \frac{P^{\mu} - K^{\mu} }{2}$ and $J^{4-1}=-J^{-14} = D$. Then these operators follow the Lorentz algebra (up to some signs that might be wrong in my notes and I will not try to verify now)

$$ [J^{ab}, J^{cd}] = i( g^{bc}J^{ad} - g^{ac}J^{bd} - g^{bd}J^{ac} + g^{ad}J^{bc} ) $$

if and only iff $g^{-1-1}=g^{00}=1$, $g^{11}=g^{22}+g^{33}=-1$, $g^{ab}=0$ for $a\neq b$. In other words the operators $J^{ab}$ create the Lorentz algebra in a spacetime of two time dimensions and 4 space dimensions. Thus $SO(4,2)$ . Now, note that this is the symmetry group of $AdS$ embedded into 2+4 dimensions as a hypersurface $g_{ab}=X^aX^b=R^2$ where $R$ is thr $AdS_5$ radius. If you try to find the isometries of $AdS_5$ via the very lengthy computation of Killing vectors etc, you will find $SO(4,2)$ and this should give you the information you want via $AdS$/CFT . More generally, $AdS_D$ has the symmetry group $SO(p+1,q+1)$ for $D=p+q$ which isomorphically matches with the one of $\mathbb{R}^{p,q}$. The $U(1)_R$ part is easier to understand. Holographically it is what survives after the Killing spinor equations come in action. It is the $R$ symmetry group for $\mathcal{N}=1$.