To solve the problem, let's examine each proposition step by step:

Proposition A:

Given $\overrightarrow{a}=(2, t)$ and $\overrightarrow{b}=(t-2, 4)$. For $\overrightarrow{a} \parallel \overrightarrow{b}$, the cross product of their components must equal zero, which gives us the equation:
$$2 \cdot 4 - (t-2) \cdot t = 0$$
Simplifying the equation:
$$8 - (t^2 - 2t) = 0$$
$$t^2 - 2t - 8 = 0$$
Solving this quadratic equation, we find:
$$t = -2 \text{ or } 4$$
Since $t$ can be $-2$ or $4$, proposition A is incorrect.

Proposition B:

The dot product of vectors is not distributive over vector multiplication in the manner described, meaning:
$$({\overrightarrow{a} \cdot \overrightarrow{b}})\overrightarrow{c} \neq \overrightarrow{a}({\overrightarrow{b} \cdot \overrightarrow{c}})$$
in general, unless $\overrightarrow{a}$ and $\overrightarrow{c}$ are collinear. Therefore, proposition B is incorrect.

Proposition C:

Given $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors and $〈{\overrightarrow{a}, \overrightarrow{b}}〉=\frac{π}{3}$, the dot product $\overrightarrow{a} \cdot \overrightarrow{b} = \cos(\frac{\pi}{3}) = \frac{1}{2}$. For $(2\overrightarrow{a}-\overrightarrow{b}) \perp \overrightarrow{b}$, we check the dot product:
$$(2\overrightarrow{a}-\overrightarrow{b}) \cdot \overrightarrow{b} = 2\overrightarrow{a} \cdot \overrightarrow{b} - \overrightarrow{b} \cdot \overrightarrow{b}$$
Substituting the given values:
$$= 2 \times \frac{1}{2} - 1 = 0$$
Since the dot product is zero, $(2\overrightarrow{a}-\overrightarrow{b}) \perp \overrightarrow{b}$, making proposition C correct.

Proposition D:

Given $\overrightarrow{a}=(-2, -1)$ and $\overrightarrow{b}=(1, 2)$, the projection of $\overrightarrow{a}$ onto $\overrightarrow{b}$ is calculated as:
$$\frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{b}|^2} \overrightarrow{b}$$
Calculating the dot product and magnitude:
$$\overrightarrow{a} \cdot \overrightarrow{b} = (-2) \cdot 1 + (-1) \cdot 2 = -4$$
$$|\overrightarrow{b}|^2 = 1^2 + 2^2 = 5$$
The projection vector is:
$$\frac{-4}{5} \cdot \frac{\overrightarrow{b}}{|\overrightarrow{b}|} = \frac{-4}{5}(1, 2) = \left(-\frac{4}{5}, -\frac{8}{5}\right)$$
Proposition D is correct.

Therefore, the correct propositions are C and D, making the final answer $\boxed{CD}$.