Let $G$ be a Lie group and $X$ a tangent vector in $T_eG$, we define the vector field $$\bar{X}_{x}=\widetilde{X}_{x}-\check{X}_{x}$$
where $\widetilde{X}_{x}=\left(L_{x}\right)_{*} X ~~~~\text {and } ~~~~\check{X}_{x}=\left(R_{x}\right)_{*} X$.
$L_x$ the left multiplication by $x$ and $R_x$ the right multiplication by $x$.
I tried to calculate the differential of the map at the identity $e$
$$X_{R}(x)=G \rightarrow T_{e} G \quad, \quad X_{R}(x)=\left(R_{x^{-1}}\right)_{*} \bar{X}_{x}$$
But in vain, please help.
In order to differentiate a tangent vector field on a differential manifold you need a way to compare vectors at nearby tangent spaces. If you naively take a co-ordinate system and differentiate as you would in Euclidean space then the resulting bilinear map on tangent spaces will depend on the co-ordinate system you picked!
So for your question to make sense you need to specify a connection.
For example a common situation is where you have a metric on your manifold and you can take the Levi-Civita connection. For a compact Lie group you could use the Haar measure to define a bi-invariant metric - but I do not think this is unique (perhaps someone will correct this in the comments).
In any case, I will answer your question in the case where the connection comes from right translating vectors back to the tangent space at the identity, differentiating, and then right translating back to $x$:$$d_Y\bar{X}_x=R_x(D_Y(R_x^{-1}\bar{X}_x))$$
Here $R_x^{-1}\bar{X}_x$ is just a function on $G$, taking values in the fixed vector space $T_e(G)$ so we may take the ordinary differential $D_Y$, along a tangent vector $Y\in T_e(G)$.
Now $$R_x^{-1}\bar{X}_x=R_x^{-1}(\widetilde{X}_x-\check{X}_x)={\rm Ad}_x(X)-X$$
Here ${\rm Ad}_x$ is just the standard name for the action $R_x^{-1}L_x$ on $T_e(G)$.
As $X$ is a constant we have:$$ d_Y\bar{X}_x =R_x(D_Y({\rm Ad}_x(X)))= R_x[Y,X] $$ Here $[Y,X]$ denotes the Lie bracket associated to the Lie group $G$. One definition is simply $[Y,X]=D_Y({\rm Ad}_x(X)))$ which gives the above result immediately. On the other hand, if you define $[Y,X]=[L_xY,L_xX]_{x=e}$ then it is a standard result that $[Y,X]=D_Y({\rm Ad}_x(X)))$ (e.g. see https://en.wikipedia.org/wiki/Adjoint_representation#Derivative_of_Ad)
Here $[L_xY,L_xX]$ denotes the Lie bracket of vector fields, which is defined on any differential manifold.
Finally then, at $x=e$ we have $d_Y\bar{X}_x = R_e[Y,X]=[Y,X]$.