User twistor59 has addressed the part regarding the "generator" terminology, but let me give a bit more detail on the second part of the question. I'm going to restrict the discussion to matrix Lie groups for simplicity.
Some background.
Given a Lie group $G$ with Lie algebra $\mathfrak g$, there exist two mappings $\mathrm{Ad}$ and $\mathrm{ad}$, both are called "adjoint." In particular for all $g\in G$ and for all $X,Y\in\mathfrak g$, we define $\mathrm {Ad}_g:\mathfrak g\to \mathfrak g$ and $\mathrm{ad}_X$ by
$$
\mathrm{Ad}_g(X) = gX g^{-1}, \qquad \mathrm{ad}_X(Y) = [X,Y]
$$
The mapping $\mathrm{Ad}$ which takes an element $g\in G$ and maps it to $\mathrm{Ad}_g$ is a representation of $G$ acting on $\mathfrak g$, while the mapping $\mathrm{ad}$ which takes an element $X\in \mathfrak g$ and maps it to $\mathrm{ad}_X$ is a representation of $\mathfrak g$ acting on itself.
In other words, $\mathrm{Ad}$ is a Lie group representation while $\mathrm{ad}$ is a Lie algebra representation, but they both act on the Lie algebra which is a vector space.
Aside.
In response to user Christoph's comment below. Note that if we define the conjugation operation $\mathrm{conj}$ by
$$
\mathrm{conj}_g(h) = g h g^{-1}
$$
Then for matrix Lie groups (which I initially stated I was restricting the discussion to for simplicity) we have
$$
\frac{d}{dt}\Big|_{t=0}\mathrm{conj}_g(e^{tX}) =\mathrm{Ad}_g X
$$
Addressing the question.
Having said all of this, in my experience (in high energy theory), physicists usually are referring to $\mathrm{ad}$, the Lie algebra representation. In fact, you'll often see it written in physics texts that
generators $T_a$ of the Lie algebra furnish the adjoint representation provided $(T_a)_b^{\phantom bc} = f_{ab}^{\phantom{ab}c}$.
where the $f$'s are the structure constants of the Lie algebra with respect to the basis $T_a$;
$$
[T_a,T_b] = f_{ab}^{\phantom{ab}c} T_c
$$
But notice that
$$
\mathrm{ad}_{T_a}(T_b) = [T_a,T_b] = f_{ab}^{\phantom{ab}c} T_c
$$
which shows that the matrix representations of the generators in the Lie algebra representation $\mathrm{ad}$ precisely have entries given by the structure constants.
Addendum (May 22, 2013).
Let a Lie-algebra valued field $\phi$ on a manifold $M$ be given. If the field transforms under the representation $\mathrm{Ad}$ (which is a representation of the group acting on the algebra) then we have
$$
\phi(x)\to \mathrm{Ad}_g(\phi(x)) = g\phi(x) g^{-1}
$$
But recall that (see here) $\mathrm{Ad}$ is related to $\mathrm{ad}$ (a representation on the algebra acting on itself) as follows: Write an element of the Lie group as $g=e^X$ for some $X$ in the algebra (here we assume that $G$ is connected) then
$$
\mathrm{Ad}_g(\phi(x)) = e^{\mathrm{ad}_X}\phi(x) = \phi(x) + \mathrm{ad}_X(\phi(x)) +\mathcal O(X^2)
$$
so that the corresponding "infinitesimal" transformation law is
$$
\delta\phi(x) = \mathrm{ad}_X(\phi(x))
$$
So when talking about a field transforming under the adjoint representation, $\mathrm{Ad}$ and $\mathrm{ad}$ in some sense have the same content; $\mathrm{ad}$ is the "infinitesimal" version of $\mathrm {Ad}$
This post imported from StackExchange Physics at 2014-03-22 17:13 (UCT), posted by SE-user joshphysics
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