Let's break down the solution step by step, following the given rules:

Statement A:

Given $|\overrightarrow{a}-\overrightarrow{b}|=|\overrightarrow{a}|+|\overrightarrow{b}|$, squaring both sides yields:
$$\begin{align*} {|\overrightarrow{a}-\overrightarrow{b}|}^2 &= {(|\overrightarrow{a}|+|\overrightarrow{b}|)}^2 \\ \overrightarrow{a}^2 - 2\overrightarrow{a}\cdot\overrightarrow{b} + \overrightarrow{b}^2 &= |\overrightarrow{a}|^2 + 2|\overrightarrow{a}||\overrightarrow{b}| + |\overrightarrow{b}|^2 \end{align*}$$
This simplifies to:
$$\begin{align*} -2\overrightarrow{a}\cdot\overrightarrow{b} &= 2|\overrightarrow{a}||\overrightarrow{b}| \\ \overrightarrow{a}\cdot\overrightarrow{b} &= -|\overrightarrow{a}||\overrightarrow{b}| \end{align*}$$
Since both $\overrightarrow{a}$ and $\overrightarrow{b}$ are non-zero vectors, it follows that:
$$\begin{align*} \cos\langle\overrightarrow{a},\overrightarrow{b}\rangle &= \frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|} = -1 \end{align*}$$
As $0 \leq \langle\overrightarrow{a},\overrightarrow{b}\rangle \leq \pi$, we have $\langle\overrightarrow{a},\overrightarrow{b}\rangle = \pi$, indicating that $\overrightarrow{a}$ and $\overrightarrow{b}$ are collinear and in opposite directions. Thus, statement A is correct.

Statement B:

Given $\overrightarrow{a}\parallel\overrightarrow{b}$ and $\overrightarrow{b}\parallel\overrightarrow{c}$, there exist non-zero real numbers $\lambda, \mu \in \mathbb{R}$, such that:
$$\begin{align*} \overrightarrow{b} &= \lambda\overrightarrow{a} \\ \overrightarrow{c} &= \mu\overrightarrow{b} \end{align*}$$
Hence, $\overrightarrow{c} = \mu\lambda\overrightarrow{a}$, implying $\overrightarrow{c}\parallel\overrightarrow{a}$. Thus, statement B is correct.

Statement C:

Given $\overrightarrow{a}\cdot\overrightarrow{c} = \overrightarrow{b}\cdot\overrightarrow{c}$, it follows that:
$$\begin{align*} \overrightarrow{a}\cdot\overrightarrow{c} - \overrightarrow{b}\cdot\overrightarrow{c} &= 0 \\ (\overrightarrow{a}-\overrightarrow{b})\cdot\overrightarrow{c} &= 0 \end{align*}$$
This implies $\overrightarrow{a}-\overrightarrow{b} = \overrightarrow{0}$ or $(\overrightarrow{a}-\overrightarrow{b})\perp\overrightarrow{c}$, meaning $\overrightarrow{a} = \overrightarrow{b}$ or $(\overrightarrow{a}-\overrightarrow{b})\perp\overrightarrow{c}$. Thus, statement C is incorrect because it does not account for the possibility of $(\overrightarrow{a}-\overrightarrow{b})\perp\overrightarrow{c}$.

Statement D:

Given $|\overrightarrow{a}+\overrightarrow{b}| = |\overrightarrow{a}-\overrightarrow{b}|$, squaring both sides gives:
$$\begin{align*} {|\overrightarrow{a}+\overrightarrow{b}|}^2 &= {|\overrightarrow{a}-\overrightarrow{b}|}^2 \\ \overrightarrow{a}^2 + 2\overrightarrow{a}\cdot\overrightarrow{b} + \overrightarrow{b}^2 &= \overrightarrow{a}^2 - 2\overrightarrow{a}\cdot\overrightarrow{b} + \overrightarrow{b}^2 \end{align*}$$
This simplifies to $\overrightarrow{a}\cdot\overrightarrow{b} = 0$, indicating that $\overrightarrow{a}\perp\overrightarrow{b}$. Thus, statement D is correct.

Therefore, the correct statements are: $\boxed{\text{A, B, and D}}$.