Indeed the Linear Least Squares uses the Covariance (Which resembles Cross Correlation) and the Variance (Which resembles the Auto Correlation) for parameter estimation.
Let's see that using simple example - Linear Function fitting.
Assume our model is given by (Simple Polynomial Model of Order 2):
$$ {y}_{i} = a {x}_{i} + b, \; i = 1, 2, \cdots, n $$
In this model we assume knowing $ \left\{ \left( {x}_{i}, {y}_{i} \right) \right\}_{i = 1}^{n} $ and we're after estimating $ a, b $.
So, we could build this in the form of Ordinary Least Squares:
$$ \boldsymbol{y} = X \boldsymbol{\theta} \Rightarrow \hat{\boldsymbol{\theta}} = {\left( {X}^{T} X \right)}^{-1} {X}^{T} y $$
Let's examine it using basic Math.
Estimating $ b $ using the Least Squared Error:
$$ \hat{b} = \arg \min_{b} \frac{1}{2} \sum_{i = 1}^{n} {\left( {y}_{i} - a {x}_{i} + b \right)}^{2} $$
Taking the Derivative with respect to $ b $ will yield:
$$ \hat{b} = a \sum_{i = 1}^{n} {x}_{i} - \sum_{i = 1}^{n} {y}_{i} = a \bar{x} - \bar{y} $$
Where $ \bar{x} = \sum_{i = 1}^{n} {x}_{i} $ and $ \bar{y} = \sum_{i = 1}^{n} {y}_{i} $.
It suggests that the constant $ a $ is set such that the line must go through the mean of $ \left\{ {x}_{i} \right\}_{i = 1}^{n} $ and $ \left\{ {y}_{i} \right\}_{i = 1}^{n} $. Which indeed makes sense, as this is the mean of the 2D data.
Estimating $ a $ using the Least Squared Error:
$$ \hat{a} = \arg \min_{a} \frac{1}{2} \sum_{i = 1}^{n} {\left( {y}_{i} - a {x}_{i} + b \right)}^{2} $$
Taking the Derivative with respect to $ a $ will yield:
$$\begin{aligned}
\frac{\mathrm{d} \frac{1}{2} \sum_{i = 1}^{n} {\left( {y}_{i} - a {x}_{i} + b \right)}^{2} }{\mathrm{d} a} & = -\sum_{i = 1}^{n} {x}_{i} \left( {y}_{i} - a {x}_{i} + b \right) && \text{} \\
& = -\sum_{i = 1}^{n} {x}_{i} {y}_{i} + a \sum_{i = 1}^{n} {x}_{i}^{2} -b \sum_{i = 1}^{n} {x}_{i} && \text{} \\
& = -\sum_{i = 1}^{n} {x}_{i} {y}_{i} + a \sum_{i = 1}^{n} {x}_{i}^{2} - \left( a \bar{x} - \bar{y} \right) \sum_{i = 1}^{n} {x}_{i} && \text{Inserting estimation of $ b $} \\
& = -\sum_{i = 1}^{n} {x}_{i} {y}_{i} + a \sum_{i = 1}^{n} {x}_{i}^{2} - a \sum_{i = 1}^{n} {x}_{i} \bar{x} + \sum_{i = 1}^{n} {x}_{i} \bar{y} && \text{} \\
\Rightarrow a & = \frac{\sum_{i = 1}^{n} {x}_{i} {y}_{i} - \sum_{i = 1}^{n} {x}_{i} \bar{y}}{\sum_{i = 1}^{n} {x}_{i}^{2} - \sum_{i = 1}^{n} {x}_{i} \bar{x}}
\end{aligned}$$
It looks like we're close but not right there.
The trick is noting these equations:
$$ \sum_{i = 1}^{n} \left( \bar{x} \bar{y} - {y}_{i} \bar{x} \right) = 0 , \; \sum_{i = 1}^{n} \left( \bar{x}^{2} - {x}_{i} \bar{x} \right) = 0 $$
So now:
$$\begin{aligned}
a & = \frac{\sum_{i = 1}^{n} {x}_{i} {y}_{i} - \sum_{i = 1}^{n} {x}_{i} \bar{y}}{\sum_{i = 1}^{n} {x}_{i}^{2} - \sum_{i = 1}^{n} {x}_{i} \bar{x}} && \text{} \\
& = \frac{\sum_{i = 1}^{n} \left( {x}_{i} {y}_{i} - {x}_{i} \bar{y} \right) + \sum_{i = 1}^{n} \left( \bar{x} \bar{y} - {y}_{i} \bar{x} \right)}{\sum_{i = 1}^{n} \left( {x}_{i}^{2} - {x}_{i} \bar{x} \right) + \sum_{i = 1}^{n} \left( \bar{x}^{2} - {x}_{i} \bar{x} \right)} && \text{} \\
& = \frac{ \sum_{i = 1}^{n} \left( {x}_{i} - \bar{x} \right) \left( {y}_{i} - \bar{y} \right) }{ \sum_{i = 1}^{n} \left( {x}_{i} - \bar{x} \right) \left( {x}_{i} - \bar{x} \right) } && \text{} \\
& = \frac{ \frac{1}{n} \sum_{i = 1}^{n} \left( {x}_{i} - \bar{x} \right) \left( {y}_{i} - \bar{y} \right) }{ \frac{1}{n} \sum_{i = 1}^{n} \left( {x}_{i} - \bar{x} \right) \left( {x}_{i} - \bar{x} \right) } && \text{} \\
& = \frac{ \operatorname{Cov} \left( x, y \right) }{ \operatorname{Var} \left( x \right) }
\end{aligned}$$
Where we have the sample Covariance and the Sample Variance of the data.
If you're after an intuition why is that, the answer is simple. If wee look at the problem from a probabilistic point of a view we have a linear model. Hence estimating its parameters should be done using the 1st and 2nd moments of the data. This is also matches the Gaussian Case where those moments define the whole distribution.
P. S.
Found also this great answer https://math.stackexchange.com/a/717552/33 (Which is similar).