In the explanation of $\vec A\cdot\vec B$, why it's true that the $\oplus$ for vector become $+$ for scalar? Edit note: I've adopt an advice to use $\oplus$, which is equivalent to $+$ for vector. And I do distinguish it from that because a scalar, in normal case, can't be added to a vector.(special case)
Both $\hat i$ and $\hat j$ are unit vectors. Assume that:
$$\vec A = A_{x} \hat{i} \oplus A_{y} \hat{j}\\
\vec B = B_{x} \hat{i} \oplus B_{y} \hat{j}\\$$
One of my book computes the dot product like this:
$$Since \quad \hat i \cdot \hat i = \hat j \cdot \hat j = 1 \quad and \quad \hat i \cdot \hat j = 0\\
\begin{align}
\vec A \cdot \vec B &=(A_{x} \hat{i} \oplus A_{y} \hat{j}) \cdot (B_{x} \hat{i} \oplus B_{y} \hat{j})\\
&= A_{x} \hat i \cdot B_{x} \hat i + A_{x} \hat i \cdot B_{y} \hat j\\
&\quad + A_{y}\hat j\cdot B_{x}\hat i+A_{y}\hat j\cdot B_{y}\hat j\\
&= A_{x} B_{x} + 0\\
&\quad + 0 + A_{y} B_{y}\\
&= A_{x} B_{x} + A_{y} B_{y}\\
\end{align}$$
Why the $\oplus$, which is for vector, in the parentheses become the $+$, which is for scalar? Is this related to axioms of vector space?

Edit2: Definition from my book:
We define $\vec A \cdot \vec B$ to be the magnitude of $\vec A$ multiplied by the component of $\vec B$ in the direction of $\vec A$. Expressed as an equation,

$\vec A \cdot \vec B = AB\cos{\phi} = |\vec A||\vec B|\cos{\phi}$

As above, expand the dot product for the two parentheses, and some terms become $0$,

$\vec A \cdot \vec B = A_{x}B_{x} + A_{y}B_{y} + A_{z}B_{z}$

 A: If I understand the question correctly, you are asking about the following identity, which expresses the distributivity of the scalar product over vector addition:
$$\vec{v} \cdot (\vec{w} \oplus \vec{u}) = (\vec{v}\cdot\vec{w})+(\vec{v}\cdot\vec{u})$$
In the above equation, the operator $\oplus$ denotes vector addition, while the operator $+$ denotes scalar addition.
The reason the operator needs to change from vector addition to scalar addition is because the quantities in parentheses on the right-hand side of the equation are not vectors.  They are scalars.  Indeed if you tried to write $\oplus$ on the right-hand side the resulting expression would not make sense.
As for why it's true, in order to answer that question we would need to know how your book defines the dot product or inner product.  In an abstract vector space, an inner product is by definition a pairing that takes two vectors and returns a scalar, and that satisfies various properties, of which the identity above is one.  In more elementary treatments, the dot product may be specified by simply defining the dot products of the basis vectors $\hat{i}$ and $\hat{j}$ and extending that definition to the entire space via the distributive property above; in such a development, the distributive property may or may not be stated explicitly, but it really should be.
A: There are two equivalent ways to define the dot product. One is with the formula that sums the products of the coordinates. The other starts from the prescribed values for the basis vectors $i$ and $j$ and the assumption that the dot product will be linear in each of its arguments separately.
If necessary, I can flesh this answer out with formulas, but the words carry the idea.
A: The dot product is a bilinear product, that is, for any $\alpha,\beta\in\mathbb R$ and any $u,v,w \in V$ where $V$ is the vector space, you have
$$(\alpha u + \beta v)\cdot w = \alpha (u\cdot w) + \beta (v\cdot w)$$
and analogously for the second argument. Note that this is part of the definition of the dot product. Also note that on the left hand side we've got an addition of vectors, while on the right hand side we've got an addition of scalars.
Using this, we find
\begin{align}
a\cdot b &= (a_x\hat i + a_y\hat j)\cdot b\,\,\,\,\\
&= a_x(\hat i\cdot b) + a_y(\hat j\cdot b)\\
&= a_x(\hat i\cdot (b_x\hat i + b_y\hat j)) + a_y(\hat j\cdot (b_x\hat i + b_y\hat j))\\
&= a_x (b_x(\hat i\cdot \hat i)+b_y(\hat i\cdot \hat j))
+ a_y (b_x(\hat j\cdot \hat i)+b_y(\hat j\cdot \hat j))\\
&= a_x (1b_x + 0b_y) + a_y (0b_x + 1b_y)\\
&= a_x b_x + a_y b_y
\end{align}
A: The dot product comes from the Law of Cosines, which, in turn, can be derived using basic geometry and trigonometry.  The Law of Cosines gives us the relationship between angles (A, B, and C) and the length of sides of an arbitrary triangle:

It states: $c^2 = a^2 + b^2 - 2ab\cos{C}$
If we treat the vertex at C as the origin, and the lines extending from it as vectors, as depticted above, then we can immediately see that we can obtain information about the angle between the vectors of a and b if we know the length of c, which is just the magnitude of b subtracted from a.
So, really, we have
\begin{equation}
\left|\vec{a} - \vec{b}\right|^2 = a^2 + b^2 - 2ab\cos{C}
\end{equation}
Which can be rewritten as
\begin{equation}
\left(a_x - b_x\right)^2 + \left(a_y - b_y\right)^2 = a_x^2 + a_y^2 + b_x^2 + b_y^2 - 2ab\cos{C}
\end{equation}
We are then left with
\begin{equation}
ab\cos{C} = a_xb_x + a_yb_y
\end{equation}
This inspires us to define
\begin{equation}
\vec{a}\cdot \vec{b} = a_xb_x + a_yb_y
\end{equation}
This is known as the dot product.  It is an operation that takes two vectors, and produces a scalar and tells us what the angle between two vectors is.
Going back to your question, notice that if a and b are perpendicular, for example, if $\vec{a} = \hat{i}$ and $\vec{b} = \hat{j}$, then the dot product is 0.  Likewise, if they are the same vector, then the dot product just becomes its squared magnitude.  The last observation to make is that the dot product distributes over addition of vectors so that
\begin{equation}
\left(\vec{a} + \vec{b}\right)\cdot \vec{c} = \vec{a}\cdot\vec{c} + \vec{b}\cdot \vec{c}
\end{equation}
This enables us, then, to define dot product in a different way, without explicitly relying on defining it in terms of components, which is what your book does.  In fact, this is the first hint that vectors exhibit a rich algebra, which enables us to encapsulate their geometric properties without directly appealing to vector components.
