"Differentiation in the direction of $X$" defines a Lie algebra representation Let $G$ be a real Lie group with Lie algebra $\mathfrak g$.  We have the $\mathbb R$-algebra $\mathscr C^{\infty}(G)$ of smooth functions $G \rightarrow \mathbb R$.  I have heard that $X \mapsto dX$ defines a Lie algebra homomorphism $\mathfrak g \rightarrow \operatorname{End}_{\mathbb R} \mathscr C^{\infty}(G)$, i.e. a representation of $\mathfrak g$.
Here $dX$ is defined to be "differentiation in the direction $X$."  What does this mean?
 A: One would like to mimic the usual derivative (of functions $\Bbb R\to\Bbb R$) that is $$\lim_{t\to 0}{f(x+t)-f(x)\over t}$$
Since the output of $f$ is still a real number in the present case ($f:G\to \Bbb R$), the process of subtracting values of $f$ and dividing them by some $t\neq 0$ still makes sense. What doesn't make sense yet is $x+t$. You can't add an element of $G$ with a real number. 
Part of the reason is that a Lie group may have a dimension greater than $1$, meaning that the variable has more freedom than a single real number can encode. Therefore, we are going to use the idea of directional derivatives from the context of multivariable calculus. A direction, in our Lie context, is an element of the Lie algebra, thought of as the tangent space to the Lie group. 
Finally, given a vector $X\in \mathfrak{g}$ and a real number $t$, we need a way to move some $g\in G$ along $tX$. How does one move an element of the group in a direction indicated by an element of the algebra? This is exactly what the exponential map does. 
Therefore, assume given $f\in\mathcal{C}^\infty(G)$. Then $d_Xf\in\mathcal{C}^\infty(G)$ is defined by $$d_Xf(g)=\lim_{t\to 0}{f(g\cdot exp(tX))-f(g)\over t}$$
More info in these notes. You can also have a look at the Wikipedia on Lie derivatives - in particular, the introduction contains the equation that lies behind the phrasing "Lie algebra homomorphism".
