To find the coefficient of the $x^5$ term in the expansion of $\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)\left(x-6\right)$, we observe that when we choose one constant from one of the six parentheses to multiply with $x$ from the other five parentheses, we are essentially looking for a way to pick the constants that will form the coefficient of $x^5$. Each term gives us a choice between multiplying by $x$ or by its constant (i.e., $-1$, $-2$, $-3$, $-4$, $-5$, $-6$).

Since we need the coefficient of $x^5$, we will select the constant from one term and $x$ from the other five. This means we are summing all the ways to select a single constant term to be itself while the others contribute their $x$ term. The constants are directly added together because choosing any one of them to not contribute an $x$ results in that constant being part of the coefficient for $x^5$.

Following this logic, the sum of all the constants is:
$$-1 + (-2) + (-3) + (-4) + (-5) + (-6)$$

which equals:
$$-1 - 2 - 3 - 4 - 5 - 6 = -21$$

Therefore, the coefficient of the term containing $x^5$ in the given expansion is $\boxed{-21}$.