I have difficulty understanding functions forming vector space. I have knowledge of basic linear algebra, so I can understand the finite vector space as linear combinations of vectors of $R^n$. 
However, when it comes function as vector and functions form a vector space, I just cannot go over. 
Below are two pieces of definitions I hardly got the essences. Could you help to explain how should I understand it. I am preparing the fundamentals so that I can understand wavelet better. 
My wrong idea is:
Try set the domain of a function fixed, say $R$, then the image of the function is a infinite-dimension vector. 


 A: This is simply applying a vector space structure to functions from a given set to $\mathbb{C}$. For instance, take $A = \mathbb{R}$. The set of all complex-valued functions on $\mathbb{R}$ form a vector space where scalar multiplication and vector addition are defined as given there. This is simply showing how we can make vector spaces out of functions.
If you have seen the notation $T \in \mathcal{L}(V,W)$, this might be a little easier to grasp. That is read as "The function $T$ is in the set of all linear transformation from $V$ to $W$", where $\mathcal{L}(V,W)$ is the set of all such transformations. This set has a natural vector space structure on it, namely the one given in the snippet you posted: 
For $\alpha \in F$ (the underlying field of both $V$ and $W$) define $\alpha T$ as $v \mapsto \alpha T(v)$ and $S+T$ as $v \mapsto S(v) + T(v)$. This gives us a vector space structure, and the objects (vectors) in this vector space are linear transformations (i.e., functions). 
Does that help your understanding?
A: All that it means to be a vector space is that you can add and subtract vectors, and scale them by elements of your field. This is clearly true about functions! What makes this vector space different than, say, $\mathbb{R}^n$ is that it doesn't have a standard basis to relate a given vector to.
Actually if $A$ is finite there is a reasonable basis. Let $A = \{a_1,\ldots,a_n\}$, then this vector space is the same as $\mathbb{R}^n$, your more familiar example. You can think of each element $a\in A$ as a basis vector and the the value $f(a)$ is the component in that "direction".
More precisely, if you take the functions $f_1,\ldots,f_n$ where $f_i(a_j) = \delta_{ij}$ then these form a basis for the space, and any function $f$ decomposes as $f = \sum_{i=1}^n f(a_i)f_i$.
[Truthfully the two vector spaces just described are dual. If you have seen dual spaces before then this function space is naturally isomorphic to the dual space to the free vector space on $A$.]
The big difference when $A$ is infinite is that these vectors $\{f_i\}$ do not form a basis. Instead, they span a proper subspace consisting of functions which are non-zero on only finitely many values $a \in A$. If we take a general function and try to write $f = \sum_{i\in I}f(a_i)f_i$ we will not get a finite sum, and thus we have left the theory of vector spaces.
It is possible to work with infinite sums of vectors, but you need to add a topology to do so. This is done especially in Hilbert spaces.
The Hilbert space $L^2[0,1]$ possesses a particularly nice (orthonormal) basis, which is $\{e^{2\pi i nx}\}_{n\in\mathbb{Z}}$. By computing a function's Fourier coefficients you can relate an arbitrary vector to this basis as an (infinite) linear combination.
However, in a general function space such as yours there is no nice basis to compare vectors to, and all you need to understand about this example is that you can add, subtract and scale functions, thus they form a vector space.
A: Try not to think the way you are thinking now, you've probably focused too much on the input space and output space of a function. Try to think of a function as an object, just like a vector. And some answers from this question might help:
Why is a function space considered to be a "vector" space when its elements are not vectors?
A: Please let me know if I'm wrong, and I will remove this answer. Here is a sketch for the associativity proof. 
$V$ we define as $\{f: f: A \to \mathbb{C}\}$.
I claim $V$ is a vector space, based on the definition here. 
Let $u, v, w \in V$. Then given an arbitrary $x$ in the corresponding field,
$$\begin{align}
((u+v)+w))(x) &= (u+v)(x)+w(x) \\
&=u(x)+v(x)+w(x) \\
&= u(x)+(v+w)(x) \\
&=(u+(v+w))(x)\text{.}\end{align}$$
Hence $((u+v)+w) = (u+(v+w))$.
