People keep mentioning the restriction on the size of a Schauder basis, but I think it's more important to emphasize that these bases are bases with respect to different spans.
For an ordinary vector space, only finite linear combinations are defined, and you can't hope for anything more. (Let's call these Hamel combinations.) In this context, you can talk about minimal sets whose Hamel combinations generate a vector space.
When your vector space has a good enough topology, you can define countable linear combinations (which we'll call Schauder combinations) and talk about sets whose Schauder combinations generate the vector space.
If you take a Schauder basis, you can still use it as a Hamel basis and look at its collection of Hamel combinations, and you should see its Schauder-span will normally be strictly larger than its Hamel-span.
This also raises the question of linear independence: when there are two types of span, you now have two types of linear independence conditions. In principle, Schauder-independence is stronger because it implies Hamel-independence of a set of basis elements.
Finally, let me swing back around to the question of the cardinality of the basis.
I don't actually think (/know) that it's absolutely necessary to have infinitely many elements in a Schauder basis. In the case where you allow finite Schauder bases, you don't actually need infinite linear combinations, and the Schauder and Hamel bases coincide. But definitely there is a difference in the infinite dimensional cases. In that sense, using the modifier "Schauder" actually becomes useful, so maybe that is why some people are convinced Schauder bases might be infinite.
And now about the limit on Schauder bases only being countable. Certainly given any space where countable sums converge, you can take a set of whatever cardinality and still consider its Schauder span (just like you could also consider its Hamel span). I know that the case of a separable space is especially useful and popular, and necessitates a countable basis, so that is probably why people tend to think of Schauder bases as countable. But I had thought uncountable Schauder bases were also used for inseparable Hilbert spaces.
"Standard coordinates" typically denotes coordinates with respect to the standard basis.
Consider the following definition.
If $\mathcal{A}=(a_1,...,a_n)$ and $\mathcal{B} = (b_1,...,b_n)$ are two bases of an $n$-dimensional
linear space $V$, then the change of basis matrix from $\mathcal{B}$
to $\mathcal{A}$ is a matrix $S_{\mathcal{B}\rightarrow \mathcal{A}}$
such that:
$$[f]_{\mathcal{A}} = S[f]_{\mathcal{B}}, \forall f \in V$$
Note that:
$$S_{\mathcal{B}\rightarrow \mathcal{A}} = \left( \begin{array}{ccc} | & & |\\ [b_1]_\mathcal{A} & ... & [b_n]_\mathcal{A}\\ | & & |\end{array} \right)$$
So coming back to your question (in the comments):
if $\mathcal{B} = \{\left( \begin{array}{ccc}
6 \\
-5 \end{array} \right),\left( \begin{array}{ccc}
-7 \\
4 \end{array} \right)\}$ is a basis of $\mathbb{R}^2$, then a matrix that converts from $\mathcal{B}$ coordinates to standard coordinates $\mathcal{E} = \{\left( \begin{array}{ccc}
1 \\
0 \end{array} \right),\left( \begin{array}{ccc}
0 \\
1 \end{array} \right)\}$ in $\mathbb{R}^2$ is:
$$S_{\mathcal{B}\rightarrow \mathcal{E}} = \left( \begin{array}{ccc}
-\frac{4}{11}& -\frac{7}{11}\\
-\frac{5}{11}& -\frac{6}{11}\end{array} \right)$$
I'll leave it to you to verify this result.
Best Answer
As other answers have already pointed out, the notion of "standard basis" only applies to the very particular $K$-vector spaces $K^n$, whose elements are $n$-tuples of elements of$~K$ (scalars). I've emphasised "are", because in any $n$-dimension $K$-vector space you can represent vectors by $n$-tuples of scalars (after having chosen an ordered basis; the scalars are the coordinates of these vectors in this basis); however in general a vector and its $n$-tuple of coordinates remain two different things.
You know what the standard basis of $K^n$ is (and it is actually an ordered basis: more than just a set of vectors, it is a list where each basis vector has its own place). The one point I would like to add is mention the property that makes this particular basis stand out among other bases, either of $K^n$ or of other spaces.
Property. Any $v\in K^n$ is equal to the $n$-tuple of coordinates of $v$ with respect to the standard basis of$~K^n$.
Clearly for such a property to hold, it is necessary that such vectors $v$ be $n$-tuples of scalars in the first place, in other words that the vector space in question be $K^n$. Moreover, the coordinates $(c_1,\ldots,c_n)$ of $v$ with respect to an ordered basis $(\mathbf e_1,\ldots,\mathbf e_n)$ are by definition the scalars such that $v=c_1\mathbf e_1+\cdots+c_n\mathbf e_n$ holds; if (with $v\in K^n$) one requires that moreover $v=(c_1,\ldots,c_n)$ as stated in the property, this requires that $$ (c_1,\ldots,c_n)=c_1\mathbf e_1+\cdots+c_n\mathbf e_n. $$ This should hold for all possible values of $c_1,\ldots,c_n\in K$. In particular one can take some $c_i=1$ and all other $c_j$ zero, and this leads to the conclusion that $\mathbf e_i\in K^n$ has to be $(0,\ldots,0,1,0,\ldots,0)$ with the nonzero component at position$~i$; this should be so for every$~i$. Once these cases are checked, all others follow by linearity. So the stated property holds for the standard basis, and also characterises the standard basis of$~K^n$.