Intuition behind the Frattini subgroup
The equivalence between the two conditions is purely formal, it holds in every category of algebraic structures.
If $u$ is a non-generator and $H$ is a maximal subgroup of $G$ (by which we mean of course a maximal proper subgroup), then $\langle H \rangle = H \neq G$, hence $\langle H,u \rangle \neq G$, which implies $H = \langle H,u \rangle$ since $H$ is maximal, and therefore $u \in H$. Hence, $u$ lies in every maximal subgroup.
If $u$ is not a non-generator, choose some $X \subseteq G$ with $\langle X \rangle \neq G$ but $\langle X,u \rangle = G$. By Zorn's Lemma there is a subgroup $H$ which is maximal with the property that it contains $X$, but does not contain $u$. In fact, $\langle X \rangle$ is such a subgroup, and if $\cal C$ is a non-empty chain of such subgroups, then one can easily check that $\cup \cal C$ is a subgroup with this property. Observe that $H$ is maximal: If $K$ is a subgroup containing $H$ properly, we must have $u \in K$ and $X \subseteq K$, hence $K=G$. Hence, $H$ is a maximal subgroup not containing $u$.
Remark: Not every subgroup of a group can be enlarged to a maximal subgroup. In fact, there are groups (such as $\mathbb{Q}$) with no maximal subgroups at all. Therefore the proof is somewhat clumsy, but it works.
More generally, if $G$ is any algebraic structure, then the intersection of all proper substructures of $G$ is called the radical of $G$, and by the proof above it coincides with the set of all non-generators of $G$. If $G$ is a group, we get the Frattini subgroup. If $G$ is a left module over a ring $R$, we get its radical, which in the particular case of $G=R$ is known as the Jacobson radical. So the Frattini subgroup is really just a special case of a more general construction, whose special cases one might be familiar with.
Probably the best way to get familiar with the Frattini subgroup is to learn some of its nice properties. It is always a characteristic subgroup. If $G$ is a finite group, then $\Phi(G)$ is nilpotent. If $G$ is a finite $p$-group, then $\Phi(G)$ is the smallest normal subgroup whose quotient is elementary abelian. In this situation, Burnside's Basis Theorem states that a subset generates $G$ if and only if its image generates the $\mathbb{F}_p$-vector space $G/\Phi(G)$, which reduces the former condition to linear algebra.
Take the cyclic group $\,G:=\langle\,x\,\rangle\,$ of order $\,p^2\,\,\,,\,\,p\,$ a prime. Then $\,\Phi(G)=\langle\,x^p\,\rangle\,$ (it's easy to check this taking $\,G\,$ as a vector space over $\,\Bbb F_p=\Bbb Z/p\Bbb Z\,$ of dimension $\,2\,$).
We know that the generators of $\,G\,$ are the elements $\,x^i\,$ , with $\,(i,p)=1\iff p\nmid i\,$ , and thus all the elements of the form $\,x^{kp}\,\,,\,\,k\in\Bbb Z\,$ , are the ones that cannot generated $\,G\,$, i.e. the non-generators.
Finally, if you know the proof of the relation between the Frattini subgroup and the set of non-generators, there you can see that an element that belongs to all the maximal subgroups of $\,G\,$ has to be a non-generator as otherwise it together with some other subset would generate the whole group without being possible to drop this element from the whole generating set, and from here one can construct a maximal subgroup that won't contain that element (Zorn Lemma's calling in the general, non-finite case)...