On the first page of these notes: if $\mu\colon G\times G\to G$ is the multiplication map on a Lie group $G$, then given a point $(a,b)\in G\times G$ and letting $R_b$ and $L_a$ denote right-multiplication by $b$ and left-multiplication by $a$, respectively, the author of the notes indicates that given $(X_a,Y_b)\in T_aG \times T_bG \simeq T_{(a,b)}(G\times G)$ and given a $C^\infty$ function $f\colon G\to\mathbb R$,
$$(X_a,Y_b)(f\circ \mu(a,b)) = X_a(f\circ R_b(a)) + Y_b(f\circ L_a(b)).$$
Why does this equality hold? We know that $\mu(a,b) = R_b(a) = L_a(b)$, but how do they split apart the $X_a$ and $Y_b$?
Elements of a direct product of vector spaces $V \times W$ are properly denoted $(v,w) \in V \times W$. So, every element can be written as $(v,w) = (v,0) + (0,w)$. By using the shorthand (abuse of notation) $v = (v,0)$ and $w = (0,w)$, we write $(v,w) = v + w$.
In this way, an element $(X_a, Y_b) \in T_aM \times T_bN$ can be written as $X_a + Y_b$, where it is understood that $X_a$ will act on functions $f \in C^\infty(M)$ and $X_b$ will act on functions $g \in C^\infty(N)$.
In your case, $M = N = G$, so that $$(X_a, Y_b)(f(ab)) = X_a(f(ab)) + Y_b(f(ab)),$$ with the caveat that we are regarding the expression $f(ab)$ as three different things, depending on its placement:
Edit: If one wants to be precise (possibly pedantic), one can let $\pi_1, \pi_2 \colon G \times G \to G$ be the projection maps and write \begin{align*} X_a(f \circ R_b) & = [(\pi_1)_*(X_a,Y_b)](f \circ R_b) = (X_a, Y_b)(f \circ R_b \circ \pi_1) \\ Y_b(f \circ L_a) & = [(\pi_2)_*(X_a,Y_b)](f \circ L_a) = (X_a, Y_b)(f \circ L_a \circ \pi_2). \end{align*}