help understanding a paragraph from Linear Algebra Done Right

It's actually simpler but much more subtle than that.

Let $\mathbb R^{\{1, ..., n\}}$ be the set of all functions from $\{1, 2, ..., n\} \rightarrow \mathbb R$.

For example if $f(x) = x^2 - 3$ then $f\in \mathbb R^{\{1, ..., n\}}$.

But $f$ can be thought of as an n-tuple. If, for instance, $n= 3$.

$f = (-2, 1, 6) \in \mathbb R^3$

If $f(x) = e^x$ then $f \in \mathbb R^{\{1, 2, 3\}} = (e, e^2, e^3) \in \mathbb R^3$.

In other words: Every 3-tuple $(x_1, x_2, x_3)$ in the vector space $\mathbb R^3$ can be thought as a function $f:\{1, 2, 3\} \rightarrow \mathbb R$ where $f(1) = x_1; f(2) = x_2; f(3) = x_3)$. The set of functions and the set of n-tuples are the same thing.

I think I should just assure you that sequences $x_n$ do make up a vector space, they can be added termwise and multiplied by a constant.

A less obvious example: Let us take a sequence of rational numbers $x_n$ such that $$ x_{n+2} = x_{n+1} + x_n. $$ This is a vector space over $\mathbb Q,$ and is of dimension exactly two. You can add them and get another, you can multiply by a rational number and get another such sequence. Not sure if you have read about bases yet, anyway, here are two basis vectors for the sequence: $$ y_1 = 1, \; \; y_2 = 1, \mbox{then} \; \; y_{n+2} = y_{n+1} + y_n, $$ $$ z_1 = 1, \; \; z_2 = 3, \mbox{then} \; \; z_{n+2} = z_{n+1} + z_n. $$ The first is the Fibonacci numbers, the second the Lucas numbers. Any sequence following the recursion is a linear combination of these.