What is ample generator of a Picard group?
First of all, the Picard group of a variety is not always monogenerated, so that the notion of "the ample generator" you are referring to surely concerns a restricted class of varieties.
Furthermore, an ample line bundle (or invertible sheaf) is a line bundle $L$ which satisfies any of the following properties :
For any coherent sheaf $\mathcal F$, the sheaf $\mathcal F \otimes L^{\otimes m}$ is generated by its global sections for every $m\geq m_0(\mathcal F)$.
The map $\Phi_{|L^{\otimes m}|} : X \to \mathbb P\left(H^0\left(X,L^{\otimes m}\right)\right), x \mapsto [s_0(x): \ldots:s_N(x)]$ - where $(s_i)$ is a base of the space of global sections of $L^{\otimes m}$ - induces a embedding of $X$ in some projective space for every $m\geq m_0$.
If you are working over $\mathbb C$, which seems to be the case, this is also equivalent for $L$ to admit a smooth hermitian metric $h$ whose curvature $\Theta_h(L)$ is a (striclty) positive $(1,1)$-form (see Kodaira's embedding theorem).
Then I guess that the ample generator of some Picard group is in some case one generator which besides is ample.
For example, the most basic example consists in taking the ample line bundle $\mathcal O_{\mathbb P^n}(1)$ over $\mathbb P^n$, which generates $\mathrm{Pic}(\mathbb P^n)$.
Finally, if your question is: if the Picard group of a projective variety is monogenerated, then can we choose an ample generator? Then the answer is yes because for every $m\geq 1$, $L$ is ample iff $L^{\otimes m}$ is ample.
Edit: I forgot to mention that in this case, the unicity of "the" ample generator is clear: indeed, if $L$ and $L^{-1}$ admit non trivial sections (say $s$ and $t$) then $st$ is a non-zero section of $\mathcal O_X$ thus is constant, so that $s$ and $t$ are both non-vanishing sections, which implies that $L$ is trivial. You can apply this to $L^{\otimes m}$ to get the unicity property.
In terms of why these might be important, having an ample line bundle is equivalent to your object of study being projective. This is since, as in Henri's answer, ample implies that there in an embedding into $\mathbb{P}H^0(X,L^{\otimes m})$ for some $m$, while conversely pulling back $\mathcal{O}(1)$ from an embedding into projective space yields an ample line bundle. Being projective gives it many nice properties.