I understand your pain in self teaching as I am doing the same. I assume you know trignometry. Imagine two vectors as below and $\theta$ as angle between them
$$
\vec{a} = a_1\hat{i} + a_2\hat{j} \\
\vec{b} = b_1\hat{i} + b_2\hat{j} \tag{1}
$$
Using law of cosines it can be proven that,
$$
\text{cos }\theta = \dfrac{\vec{a}\bullet\vec{b}}{\lVert a \rVert \lVert b \rVert} \tag{2}
$$
where
$$
\vec{a}\bullet\vec{b} = a_1b_1 + a_2b_2 \tag{3}
$$
The vector form can also be expressed in matrix multiplication as below.
$$
\vec{a} \bullet \vec{b} =
\begin{matrix}
\begin{bmatrix}
a_1 & a_2
\end{bmatrix} \\[2.8ex]
\end{matrix}
\begin{bmatrix}
b_1 \\ b_2
\end{bmatrix} =
\begin{bmatrix}
a_1 \\ a_2
\end{bmatrix} \bullet
\begin{bmatrix}
b_1 \\ b_2
\end{bmatrix} = a_1b_1 + a_2b_2 = \sum_i^2 a_i b_i \tag{4}
$$
So essentially if you have sample set $(X,Y) = \{ (x_1,y_1), (x_2,y_2) \}$, you could visualize 2D vectors constructed out of this sample set $(X,Y)$.
Let the 2D vectors be
$$
\vec{x} = x_1\hat{i} + x_2\hat{j} \\
\vec{y} = y_1\hat{i} + y_2\hat{j} \tag{5} \\
$$
$$
\text{cos }\theta = \dfrac{\vec{x}\bullet\vec{y}}{\lVert x \rVert \lVert y \rVert} \tag{6}
$$
where
$$
\vec{x}\bullet\vec{y} = x_1y_1 + x_2y_2 \tag{7}
$$
In matrix multiplication form,
$$
\vec{x} \bullet \vec{y} =
\begin{matrix}
\begin{bmatrix}
x_1 & x_2
\end{bmatrix} \\[2.8ex]
\end{matrix}
\begin{bmatrix}
y_1 \\ y_2
\end{bmatrix} =
\begin{bmatrix}
x_1 \\ x_2
\end{bmatrix} \bullet
\begin{bmatrix}
y_1 \\ y_2
\end{bmatrix} = x_1y_1 + x_2y_2 = \sum_i^2 x_i y_i \tag{8}
$$
Now if you center the sample set, that is subtract each sample point from its mean, still the law of cosine could be applied, but with now a slightly different vector.
Let
$$
\vec{x_c} = (x_1 - \overline{x})\hat{i} + (x_2 - \overline{x})\hat{j} \\
\vec{y_c} = (y_1 - \overline{y})\hat{i} + (y_2 - \overline{y})\hat{j} \tag{9}
$$
Then using same steps as above, we can show that,
$$
\text{cos }\theta = \dfrac{\vec{x_c}\bullet\vec{y_c}}{\lVert x_c \rVert \lVert y_c \rVert} \tag{10}
$$
And the dot product,
$$
\vec{x_c} \bullet \vec{y_c} =
\begin{matrix}
\begin{bmatrix}
x_1-\overline{x} & x_2 -\overline{x}
\end{bmatrix} \\[2.8ex]
\end{matrix}
\begin{bmatrix}
y_1 -\overline{y} \\ y_2 -\overline{y}
\end{bmatrix} =
\begin{bmatrix}
x_1 -\overline{x} \\ x_2 -\overline{x}
\end{bmatrix} \bullet
\begin{bmatrix}
y_1 -\overline{y} \\ y_2 -\overline{y}
\end{bmatrix} = \sum_i^2 (x_i - \overline{x}) (y_i - \overline{y}) \tag{11}
$$
and modulus of the vectors of course,
$$
\lVert x_c \rVert = \sqrt{\sum_i^2 (x_i - \overline{x})^2} \\
\lVert y_c \rVert = \sqrt{\sum_i^2 (y_i - \overline{y})^2} \tag{12} \\
$$
Thus eq.(10) becomes,
$$
\text{cos }\theta = \dfrac{\vec{x_c}\bullet\vec{y_c}}{\lVert x_c \rVert \lVert y_c \rVert} = \dfrac{\sum_i^2 (x_i - \overline{x}) (y_i - \overline{y})}{\sqrt{\sum_i^2 (x_i - \overline{x})^2}\sqrt{\sum_i^2 (y_i - \overline{y})^2}} = \dfrac{\text{cov}(X,Y)}{s_Xs_Y} = r
\tag{13}
$$
where $\text{cov}(X,Y)$ is sample covariance and $s_X,s_Y$ are unbiased sample standard deviations.
This is how the $r$ and angle between the vectors constructed out of the sample set are related. I have constructed with only sample size of 2 pairs $(x_1, y_1), (x_2,y_2)$, but this could be extended to any number of sample set size.
Thus a N sized sample set, could be imagined as 2 vectors of N dimensions and of course we cannot visualize it. But in that unimaginable N dimension, the angle $\theta$ between those "vectors" still be on a plane (2D), so the law of cosine still applies. Below is an image, where for $N=3$, we could visualize 2, 3D vectors, and you can see the angle sweeped between them is still on a 2D plane. Of course we cannot go beyond 3D, but you can get the point.
Note that, the value of cosine ranges between $\pm 1$. So when both vectors are in same direction, the $\theta$ is 0, thus cos$\theta$ = 1, maximum value indicating perfect linearity. Similarly when both vectors are in opposite direction, $\theta = 180^{\circ}$, implying cos$\theta$ = -1. When the vectors are perpendicular to each other, $\theta = 90^{\circ}$ implying cos$\theta = 0$, thus zero correlation.
