Can any function $f :\mathbb{F}^m \rightarrow \mathbb{F}^n$ over some field $\mathbb{F}$ be written as
$f(x_1,...,x_m) = \begin{align} \begin{bmatrix} f_{1}(x_1,...,x_m) \\ f_{2}(x_1,...,x_m) \\ \vdots \\ f_{n}(x_1,...,x_m) \end{bmatrix} \end{align}$
where $f_i :\mathbb{F}^m \rightarrow \mathbb{F}$ $\forall i$?
An accompanying proof would be much appreciated also.
Yes, consider $p_i$ defined on $\mathbb R^n$ by $p_i(x_1,\dots,x_n)=x_i$.Then you can define $f_i=p_i\circ f$ and the formula
$$ f(x_1,...,x_m) = \begin{align} \begin{bmatrix} f_{1}(x_1,...,x_m) \\ f_{2}(x_1,...,x_m) \\ \vdots \\ f_{n}(x_1,...,x_m) \end{bmatrix} \end{align} $$
follows immediately. The functions $p_i$, $i=1,2,\dots,n$, are called canonical projection (see this Wikipedia article).
Alternatively, for every $(x_1,\dots,x_m)$ in $\mathbb R^m$, $f(x_1,\dots,x_m)$ is an element of $\mathbb R^n$, so it can be written as $f(x_1,\dots,x_n)=\begin{align} \begin{bmatrix} f_{1} \\ f_{2} \\ \vdots \\ f_{n} \end{bmatrix} \end{align}$
but the numbers $f_1,\dots,f_n$ depends on $x_1,\dots,x_m$ so we could write $f(x_1,\dots,x_n)=\begin{align} \begin{bmatrix} f_{1} \\ f_{2} \\ \vdots \\ f_{n} \end{bmatrix} \end{align}$. No proof needed since that's the definition of $\mathbb R^n$. What should be said is that, when $(x_1,\dots,x_m)$ is fixed, the real numbers $f_1,\dots,f_n$ are uniquely determined. Therefore, to every $(x_1,\dots,x_m)$ in the domain of $f$, we associate a unique element (denoted $f_i(x_1,\dots,x_m)$) of $\mathbb R$. This means that $f_i$ is a function.