Too keep things as simple as possible, assume that $\mathbf{x} = (1,2)$ and $\mathbf{y} = (3,-5)$. You will think of $\mathbf{x},\mathbf{y}$ as "vectors", i.e. as arrows that start at the center $(0,0)$ and go onto those points.
Here is a picture:
How do we find the angle $\theta$ between these vectors? There is a formula in geometry that tells us that to do so:
$$ \cos \theta = \frac{ \mathbf{x}\cdot \mathbf{y} }{ ||\mathbf{x}|| ~ ~ ||\mathbf{y} || } $$
Now we need to explain each piece of the right-hand-side. Let us begin with $||\mathbf{x}||$. This is called the (Euclidean) norm of $\mathbf{x}$. This represents the length of the vector. The way you find it is by basically using the Pythagorean theorem, in this case,
$$ ||\mathbf{x}|| = \sqrt{ 1^2 + 2^2 } = \sqrt{5} $$
In a similar way,
$$ ||\mathbf{y}|| = \sqrt{ 3^2 + (-5)^2 } = \sqrt{ 34 } $$
But how do we find $\mathbf{x}\cdot \mathbf{y}$? This is the ``dot product''. Here is how you calculate it,
$$ \mathbf{x}\cdot \mathbf{y} = (1)(3) + (2)(-5) = 3 - 10 = -7 $$
This means that,
$$ \cos \theta = \frac{-7}{\sqrt{170}} $$
Now in what follows you do not really care about the angle $\theta$ but care more about $\cos \theta$. Maybe you remember from high-school that $\cos \theta$ is always a number between $-1$ and $1$.
Now pretend we have two sets of data $\mathbf{x} = (x_1,x_2,...,x_n)$ and $\mathbf{y} = (y_1,y_2,...,y_n)$. Let us say, for simplicity, that the average $\overline{\mathbf{x}} = 0$ and $\overline{\mathbf{y}} = 0$. So the average of the two data sets is equal to zero. As before let us calculate the $\cos \theta$ of these two data sets where $\theta$ is the angle between $\mathbf{x}$ and $\mathbf{y}$. We would get,
$$ \cos \theta = \frac{\mathbf{x}\cdot \mathbf{y}}{||\mathbf{x}|| ~~ ||\mathbf{y}||}
= \frac{x_1y_1 + x_2y_2 + ... + x_ny_n}{\sqrt{x_1^2 + ... + x_n^2} ~ ~ \sqrt{y_1^2 + ... + y_n^2}} $$
Now make the profound observation that the numerator:
$$ x_1y_1 + ... + x_ny_n $$
is exactly how you compute the covariance $\text{Cov}(\mathbf{x},\mathbf{y})$! And the denominator $\sqrt{x_1^2 + ... + x_n^2}$ is exactly how you compute the standard deviation $\text{Std}(\mathbf{x})$!
Therefore, you see that the correlation between $\mathbf{x}$ and $\mathbf{y}$ is a number between $-1$ and $+1$. Furthermore, it is equal to $+1$ exactly when the two arrows (vectors) are pointing in the same exact direction, i.e. it is a perfect straight line.
