I suspect you are getting a little ahead of yourself.
First of all, to properly understand affine spaces, one should be thoroughly comfortable with the vector spaces upon which they are built.
I'm not sure I can explain this very well, but I will try.
When we try to "arithmetize" geometry, we often do so with a coordinate system. Now, if we are doing plane geometry (dealing with triangles, lines, circles, and other curves) it doesn't matter "where we are", for example a vector is a "difference of points", often denoted as:
$\vec{PQ}$
to indicate we are "starting at the point $P$, and ending at $Q$".
If $P = (x,y)$ and $Q = (x',y')$ in a typical Euclidean coordinate system, what we really mean by $\vec{PQ}$ is the "coordinate vector" $(x'-x,y'-y)$ translated by $(x,y)$.
The thing about vector spaces (which arise naturally from coordinate systems), is that they have a "special point", the origin, which has properties no other point does. But in "actual" space (or the virtual kind we idealize with geometry), no point is any more special than any other.
So if you want to study "space geometry", coordinate systems are somewhat artificial: we have to impose a choice of origin, and "some axes to measure by" which aren't inherent in the geometrical situation.
The trade-off, of course, is that once we have DONE so, we can tag points with "number arrays", and use the number-system the arrays take values from to perform calculations (like finding the angle between two vectors, or calculating the normal vector to a plane).
Another way to see this relationship is by looking at system of linear equations, like so:
$Ax = b$, where $A$ is a matrix, and $x$ and $b$ are appropriately-dimensioned vectors.
Typically the way we do this, is to first study the system:
$Ax = 0$ (the related homogeneous system). Solving this is the same thing as finding $\text{ker }A$. Matrices are, among other things, $\Bbb Z$-module homomorphisms, with some additional properties...it turns out that for (finite-dimensional) vector spaces $U,V$ over a field $F$, that $\text{Hom}_{\ F}(U,V)$ is ALSO a vector space, and if we CHOOSE bases for $U,V$ (think "coordinate systems") we can identify a "vector-space morphism" (an $F$-linear map) with a matrix $A$, that is:
$\text{Hom}_{\ F}(U,V) \cong F^{\dim(U)\cdot\dim(V)} \cong \text{Mat}_{\ \dim(V) \times \dim(U)}(F)$.
The 2nd isomorphism is "canonical", we can "vectorize" a matrix by concatenating its columns into one long array, but the first one will depend on the choice of bases.
We then get the "general solution" (to $Ax = b$) by translating the entire null space (kernel) of $A$ by the vector $b$. In fact, people are often introduced to "affine transformations" by breaking them into two parts:
Linear part + translation part.
What we call "lines in the plane" are affine transformations, the line itself is an affine space with underlying vector space $\Bbb R$.
$L = \{(x,y) \in \Bbb R^2: y = mx + b\}$
with affine function:
$f(x) = mx + b$
which has "linear part" $g(x) = mx$, and we have an isomorphism (of vector spaces):
$\Bbb R \cong L - (0,b)$ given by: $x \leftrightarrow (x,mx)$
so a line in the plane is just "a one dimensional (real) linear subspace of the vector space $\Bbb R^2$ offset by $(0,b)$". If we call the graph of $g$ (the subspace which is the line parallel to $L$ through the origin), $K$, then $L$ is a COSET of $K$, that is to say an element of $\Bbb R^2/K$, namely $L = (0,b) + K$.
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OK, now normally, we are SO USED to the standard "unit $x$-axis vector" (1,0) and the "standard $y$-axis vector" (0,1) that they are, for all intents and purposes, invisible to us. But the geometry (which may be what interests us) should be "translation invariant". A circle should be a circle no matter what coordinates we assign to its center, and its area or how many times it intersects a given line should not depend on the coordinate system we use.
My somewhat long-winded point is, an affine plane is a generalization of $\Bbb R^2$ "without coordinates" (the thing Euclid may have had in mind when he wrote The Elements), but with an arbitrary field $K$ instead of the real numbers. And to understand what this plane is like, you ought to know what the vector space $K^2$ is like.
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Which brings me to your second question:
An (associative, unital) algebra $A$ over a commutative ring $R$ is a set $A$ such that:
1. $A$ is an $R$-module (typically, $R = F$ is often a field, so that $A$ is a vector space over $F$).
2. $A$ is a ring whose multiplication is $R$-bilinear.
3. There exists $1 \in A$ such that $1x = x = x1$ for all $x \in A$.
We can start with either the $R$-module, or the ring, as our starting point:
Given an $R$-module $A$, if there is a bilinear map $(\cdot,\cdot): A \times A \to A$ such that:
$(xy)z = x(yz)$, and $A$ has a unital element, it is easy to show that $A$ is a ring with unity.
On the other hand, given a commutative ring $R$ and a ring $A$, and a ring homomorphism: $\phi:R \to Z(A)$ (where ($Z(A)$ denote the center of $A$) we can define an $R$-module action by:
$r\cdot x = \phi(r)x$, for $r \in R, x \in A$.
Note this gives us a unity in $A$, namely $\phi(1_R)$.
Some examples of algebras which are very important:
1. The polynomial ring $F[x]$, for a field $F$.
2. The ring of $n \times n$ matrices with entries in a commutative ring $R$.
3. An extension ring of a commutative ring. This extension need not be commutative, the quaternions as an extension of the reals is the usual example.
In other words, algebras have this "it's a shaving cream/it's a dessert topping" kind of duality, and one can use either ring theory, or linear algebra to investigate them, or a mixture of both.
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Long story short, get a good working knowledge of rings and linear algebra, first. Otherwise, you're likely to very quickly become confused.
