Dot product in coordinates

I suppose you want to prove that two your definitions of dot product are the same. We start with definition of dot product as $(\vec{u}, \vec{v}) = |\vec{u}| |\vec{v}| \cos \theta$. We start with definition of dot product as $(\vec{u}, \vec{v}) = |\vec{u}| |\vec{v}| \cos \theta$ and prove that it also satisfies $(\vec{u}, \vec{v}) = x_1 x_2 + y_1 y_2$.

At first you can prove that dot product is linear: $(\vec{v_1}, \vec{v_2} + \alpha \vec{v_3}) = (\vec{v_1}, \vec{v_2}) + \alpha (\vec{v_1}, \vec{v_3})$. This is true because $(\vec{v_1}, \vec{v_2})$ is equal to the product of $|\vec{v_1}|$ and projection of $\vec{v_2}$ on $\vec{v_1}$. Projection of sum of vectors is equal to sum of projections. Hence dot product is linear.

Let $\vec{e_1}$ and $\vec{e_2}$ be vectors with coordinates $(1, 0)$ and $(0, 1)$.

After that if $\vec{v_1} = x_1 \vec{e_1} + y_1 \vec{e_2}$ and $\vec{v_2} = x_2 \vec{e_2} + y_2 \vec{e_2}$ then by linearity of dot product we have $(\vec{v_1}, \vec{v_2}) = x_1 x_2 (\vec{e_1}, \vec{e_1}) + x_1 y_2 (\vec{e_1}, \vec{e_2}) + x_2 y_1 (\vec{e_2}, \vec{e_1}) + x_2 y_2 (\vec{e_2}, \vec{e_2})$. Since $(\vec{e_1}, \vec{e_1}) = (\vec{e_2}, \vec{e_2}) = 1$ and $(\vec{e_1}, \vec{e_2}) = (\vec{e_2}, \vec{e_1}) = 0$ we have $(\vec{v_1}, \vec{v_2}) = x_1 x_2 + y_1 y_2$.

The dot product is invariant under rotations, we may therefore rotate our coordinate system so that v is along the x-axis. In this case, $v = (|v|, 0)$. Letting $w = (x,y)$ we have (using the definition of dot product in Cartesian coordinates) $v \cdot w = |v| x$. But what is $x$? Well, if you draw the picture and let $\theta$ be the angle between v and w, then we see that $\cos \theta = x/|w||$ so that $x = |w| \cos \theta$. Thus $v\cdot w = |v||w| \cos \theta$.