$f$ in $f(x)$ as a vector I might split this question into two, so the first paragraph will contain the main question.
Given a linear function $f(x,y)$, is it possible to consider $f$ as a vector? Given the relationship of the dot product, $(a,b)\cdot(c,d) = ab+cd$, it seems to me that a given function $f(x,y) = ax+by$ can be written in vector notation as $(a,b)\cdot(x,y) = \vec{f}\cdot(x,y)$, where $f \sim \vec{f}$. Is this a just a nice coincidence, or does this have any practical uses?
The part that I might split into a second question is: Does this generalize in any way to non-linear functions? It doesn't seem to be possible to use this with constants ($+k$) or higher order polynomials ($x^2+y^3$) $\textit{(If this deems the question too broad, simply ignore this.)}$
 A: Every linear function $L$ on $\mathbb{R}^n$ can be represented by a vector through the standard inner product. This is almost obvious, since
$$
L(x_1,\ldots,x_N) = \sum_{k=1}^N a_k x_k
$$
for suitable numbers $a_k$. Hence $L(x_1,\ldots,x_N)  = \langle (x_1,\ldots,x_N) \mid (a_1,\ldots, a_N) \rangle$.
This is actually a special case of a much deeper result called Riesz Representation Theorem.
A: 1) Yes. The set of all linear functions from $\mathbb{R}^2$ to $\mathbb{R}$ is called the dual vector space $(\mathbb{R}^2)^*$of $\mathbb{R}^2$. Since $\mathbb{R}^2$ is a finite-dimensional vector space, $\mathbb{R}^2$ and $(\mathbb{R}^2)^*$ are isomorphic, although there are many ways to identify them, and there is no canonical isomorphism.
2) It is possible to consider very many vector spaces $V$ of functions from $\mathbb{R}^2$ to $\mathbb{R}$. Examples: Spaces of affine functions, polynomial functions, continuous functions, bounded functions, or any functions. However, the application operation $\bullet: V \times \mathbb{R}^2 \to R$ with $f \bullet (x, y) := f(x, y)$ is not bilinear if $V$ contains any nonlinear function, so you cannot actually represent it as something bilinear (such as a scalar product).
If you however want to transform the vector $(x, y)$ nonlinearly beforehand, then you might very well be able to construct a scalar product. Consider for example the space of quadratic polynomials $Q := \{ax^2+bxy+cy^2+dx+ey+f: a, b, c, d, e, f \in \mathbb{R}\}$.
Then for $p := ax^2+bxy+cy^2+dx+ey+f \in Q$ you can actually represent $p$ as a scalar product, like this:
$$p(x, y) = \begin{pmatrix}a\\b\\c\\d\\e\\f\end{pmatrix}\cdot\begin{pmatrix}x^2\\xy\\y^2\\x\\y\\1\end{pmatrix}$$
It's actually rather easy in the finite-dimensional case. Just pick a basis $f_1, \ldots, f_n$ of your function space, then for any $f = a_1f_1 + \ldots + a_nf_n$ you can write
$$f(x) = \begin{pmatrix}a_1\\\vdots\\a_n\end{pmatrix}\cdot\begin{pmatrix}f_1(x)\\\vdots\\f_n(x)\end{pmatrix}$$
