Is any (basis or non-basis) vector in essence always a linear combination of standard basis vectors? Is it true that the vectors in the basis of any vector space are always coordinate vectors with respect to the standard basis of the (possibly larger) vector space we are working in? And by extension is it true then that any "vector" in said vector space is in essence the coordinate vector with respect to the standard basis? And further, is the standard basis always a basis where the coordinate vector with respect to the standard basis (in this case, itself), of any vector in it, is that same coordinate-vector? (Because it seems like the coordinates of all basis-vector in a non-standard basis with respect to the standard basis will never all be the same as that vector's coordinate to its own basis, which is just a vector with one entry of one and all others zero, but I'll have to prove that another day.)
For example, take bases $ \{(1,0),(0,1)\} $ and $ \{(1,1),(0,1)\} $ (brackets aren't rendering in my preview, don't know why) for $\mathbb R^2$. The 4 vectors in both bases themselves are coordinates for $\mathbb R^2$, that tell you how to scale some other basis-vectors to find what they represent. (The question is if these "other vectors" are the standard basis vectors. I think they need to exist and be the standard basis vectors because otherwise things wouldn't make sense, but it's hard to explain that.) We all know that $(1,0)$ means "go one step to the right and zero up," or "scale your first basis vector by one and your second by zero." But if your basis vectors in essence weren't measured relative to the standard basis, we would end up in an infinite regress, because you could ask "well what is the basis of your basis-vectors?" "It could as well be anything!", right?
Suppose we take the "vector" with "coordinate-vector $(5,2)$ with respect to the second basis", then we're really talking about $5(1,1)+2(0,1)=$ $(5,7)$, where $(5,7)$ says "go five right and seven up and then you end up at the essence of that vector", right?
By further extension, is it true then that an $n$-tuple (vector) in an $n$-dimensional vector space is actually an $\infty$-tuple in an $\infty$-dimensional vector space but where we just focus on the $n$ relevant components and leave out all the $0$'s?
For clarity, what I mean by "coordinate vector" is: suppose we have basis $\beta=\{b_1,\ldots,b_n\}$ for vector space $(\mathbb F,V,+)$ with $\dim V=n$. Then there exists a function $\text{co}_\beta:$ $V\rightarrow\mathbb F^n:$ $\sum_{k=1}^n\lambda_kb_k\mapsto(\lambda_1,\ldots,\lambda_n)$ that maps a vector to an n-tuple with its coordinates with respect to $\beta$.
I've found similar questions on here but not exactly the same I think. I'm sorry that I couldn't formulate my question better, I did my best. Edit: I may have sort of answered my own question here during edits, so I'm sorry if that inappropriate as it may have switched the original post from "question" status to more of a "discussion" status. The question would still be if anyone can clear this up 100%.
 A: In a general vector space, there is no "standard" / "canonical" basis.
Only in $F^n$ (as a space over F) you have such thing.

*

*Think of the space of all continuous functions, for instance.

*Or the tangent space to a sphere in a given point.

*Or all polynomial $P$ of order three such that $P(1) = 0$.

*Or a k-dimensional subspace of $F^n$,

*...

You can choose many good bases suited for your application, but not a canonical choice.
A: You say "it's hard to explain". So let's take an example to try to explain. You briefly recalled what a vector space is over a field $\mathbb F$.
Let us limit ourselves to real vector spaces $(\mathbb F=\Bbb R)$. Let's take $V=\mathcal{F}(\Bbb R,\Bbb R)$[you can add two functions and multiply a function by a scalar and all the axioms of a vector space hold]. But here no "standard basis".
Let's see what happens when we limit ourselves to the vector subspace $W$ of $V$ composed of the polynomial functions($x\mapsto \frac73x-\pi x^{942}, x\mapsto x^2+1, x\mapsto x+\sqrt2,... $ are vectors of $W$). Here we can try to talk about the basis $$(1,X,X^2,...X^n,...)$$
where, for $n \in \mathbb N, X^n$ denotes the vector $\mathbb R \to \mathbb R, x\mapsto x^n$.
For example, what you said "for clarity" applies here for the subspace $W_{n-1}(n>1)$ of polynomials of degree less than or equal to $n-1$, with $\beta=\{1,X,...X^{n-1}\}$ for example.
Consider $W_1=span(1,X)=\{f\in \mathcal{F}(\Bbb R,\mathbb R):\exists a_0,a_1 \in \Bbb R, \forall x \in \Bbb R, f(x)=a_0+a_1 x \}$
Take basis $\{1+X,X\}$. The vector whose coordinates in this second basis are $5$ and $2$, is $5(1+X)+2X=5+7X$ or more simply $x\mapsto 5+7x$.
In my view, the question is not: is this correct? The question is: what is it for?
I suggest the following illustration (there are a multitude of them):
Let $f \in W_n$ for some $n>1$. Let $\alpha \in \Bbb R$. Then there are elements $c_0,...,c_n $such that$$f(x)=c_0+c_1(x-\alpha)+...+(x-\alpha)^n$$
Let $(Y^k)_{k \in [\![0,n]\!]}$ defined by $Y^k=(X-\alpha)^k$. $(Y^k)_{k \in [\![0,n]\!]}$ is a new basis for $W_n$.
What is it for ? Note that we have $f(\alpha)=c_0$. Hence, if $f(\alpha)=0, c_0=0$ and we can write $$f(x)=(x-\alpha)h(x)$$
expression in which we can still write$$h(x)=d_1+d_2(x-\alpha)+...+d_n(x-\alpha)^{n-1}$$
And so on... The important thing is that it is useful for something in mathematics.
