How can an element of $\mathbb F^n$ be thought of as a function $\{1,2, \dots, n\} \to \mathbb F$? I have a good knowldge of abstract algebra  - stating this just in case this helps with an answer.
My linear algebra book states $$\mathbb F^n= \underbrace{\mathbb F \times \cdots \times \mathbb F}_\text{$n$-many} $$ can be thought of as $\mathbb F^{\{1,2, \dots, n\}}$ which denotes the set of functions $\{1,2, \dots, n\} \to \mathbb F$. I don't see how at all.
For instance, what function the element $(1,2) \in \mathbb{F}^2$ corresponds to in $\mathbb{F}^{\{1,2\}}$? Do the elements $1,2 \in \{1,2\}$ get mapped to $1\in \mathbb F$? 
 A: A function $f:\{1,2, \dots, n\} \to \mathbb F$ is uniquely defined by the values at the $n$ points i.e. the $n$-tuple $\big(f(1), f(2), \ldots, f(n)\big) \in \mathbb{F}^n$.
A: A function from $\{1\dots n\}$ to $\mathbb{F}$, will map the coordinates of a point in $\mathbb{F}^n$ to the coordinates of that point.
For example, the function from $\{1,2,3\}\to\mathbb{R}$ given by
$$
f(1)=3.24,f(2)=1.222,f(3)=-21.44
$$
would correspond to the point in $\mathbb{R}^3$
$$
(3.24,1.222,-21.44)
$$
A: For $v\in\mathbb{F}^n$, it's uniquely represented as a linear combination of some basis elements (for some fixed, ordered basis $\beta = \{e_1,\dots,e_n\}$):
$$v = \sum_{i = 1}^n a_ie_i$$
Now, we'll establish the isomorphism via the bijection:
$$\phi(e_i) = \delta_i,\quad \delta_i:\{1,\dots,n\}\to\mathbb{F},\quad \delta_i(j) = \begin{cases} 1 & i = j \\ 0 & \text{else}\end{cases}$$
This can easily be seen to be a bijection, but I won't go through verifying it here.
So, we have that:
$$\phi(v) = \phi\left(\sum_{i = 1}^n a_ie_i\right) = \sum_{i = 1}^n a_i\phi(e_i) = \sum_{i = 1}^n a_i\delta_i$$
This is a function $\{1,\dots,n\}\to\mathbb{F}$, that assigns the value of $a_i\in\mathbb{F}$ to input value of $i$.
A: An element $(x_{1}, \dots, x_{n}) \in \Bbb F^{n}$ can be thought of as the function $f : \{1,\dots,n\} \to \Bbb F$ defined by $f(1) = x_{1}$ (i.e., $f(1)$ is the first coordinate), $f(2) = x_{2}$ (i.e., $f(2)$ is the second coordinate), and so on.
So, in your example, $(1,2) \in \Bbb F^{2}$ can be thought of as the function $f : \{1,2\} \to \Bbb F$ defined by $f(1) = 1$ and $f(2) = 2$.  
In the $n$ case, $f(i)$ is the $i$-th coordinate.  So for the element $(3,-5,20)$, this would be the function $f: \{1,2,3\} \to \Bbb F$ defined by $f(1) =3$, $f(2) = -5$, and $f(3) = 20$.
A: In general for any set, $S$ and any index set, $I$, the set of $I$-tuples of elements of $S$ should be thought of as $S^I$, the set of functions from $I$ to $S$. If $i\in I$, $f\in S^I$, the $i$th element of $f$ is $f(i)$, usually denoted $f_i$. 
Examples:


*

*The set of sequences in $X$ is $X^\Bbb{N}$

*Ordered pairs of elements of $X$, $X\times X$ is naturally bijective to $X^2$, where we think of 2 as the set $\{0,1\}$ or $\{1,2\}$ depending on how you like to start your indices. The bijection sends the function $f:2\to X$ to the ordered pair $(f(0),f(1))$, and sends the ordered pair $(x_0,x_1)$ to the function $0\mapsto x_0,1\mapsto x_1$. 

*The previous example generalizes. As in your question, ordered $n$-tuples of elements of $X$ may be thought of as $X^n$, where we think of $n$ as either $\{0,1,\ldots,n-1\}$ or $\{1,2,\ldots,n\}$

