A module is projective iff it has a projective basis

The statement you're linking to is: A module $P$ is projective if and only if there is a family $\{x_{i}\}_{i \in I} \subset P$ and morphisms $f_{i}: P \to R$ such that for each $x \in P$ we have $x = \sum_{i \in I} f_{i}(x) x_{i}$. The last statement says three things:

1. In order for the sum to make sense we must have that for all $x$ the set $\{i\,:\,f_{i}(x) \neq 0 \}$ is finite. Or, as stated there: for all $x$ we have $f_{i}(x) = 0$ for almost all $i$.
2. The set $(x_{i})_{i \in I}$ generates $P$. In other words, the map $g: \bigoplus_{i \in I} R \to P$ sending $(r_{i})$ to $\sum_{i \in I} r_{i} x_{i}$ is an epimorphism (it suffices to take $r_{i} = f_{i}(x)$ to see that this map is onto).
3. The epimorphism $g$ splits: there is a right inverse $f: P \to \bigoplus_{i \in I} R$ of $g$, i.e. $gf = \operatorname{id}_{P}$ (this morphism $f$ is of the form $(f_{i})_{i \in I}$ with morphisms $f_{i}: P \to R$ and by the definition of a direct sum we have $f_{i}(x) \neq 0$ only for finitely many $i$).

In my opinion its just an extremely explicit way of phrasing the much more catchy "a module is projective if and only if it is a direct summand of a free module".

The statement "every element in P can be written as a finite linear combination of some elements of P.", where "some" means a finite set, just says that the module is finitely generated.

This has nothing to do with being projective.

Take for instance the $\mathbb Z$-module $\mathbb Z/2$. Here every element can be written as a multiple of $[1]$. So $\mathbb Z/2$ is finitely generated but not projective.

On the other hand the infinite direct sum $\bigoplus_{\mathbb N}\mathbb Z$ is a projective $\mathbb Z$-module which is not finitely generated.

The crucial thing in the definition of a projective basis is really that you have homomorphisms out of $P$ into the ring (regarded as module over itself).

The phrasing of your statement

a module P is projective iff every element in P can be written as a finite linear combination of some elements of P

is pretty bad, because it is not at all evident from that those some elements are fixed: what you mean, but not what you wrote, probably is

a module is projective iff there is a set $X\subseteq P$ such that every element of $P$ is a finite combination of elements of $X$.

Now, this last statement is false: you can always pick $X=P$, so its truth would imply that all modules are projective---and there are rings which have non-projective modules!