(1) Since $\overrightarrow{AB} = \overrightarrow{a} + \overrightarrow{b}$, $\overrightarrow{BC} = 2\overrightarrow{a} + 8\overrightarrow{b}$, and $\overrightarrow{CD} = 3(\overrightarrow{a} - \overrightarrow{b})$, we can find $\overrightarrow{BD}$ by adding $\overrightarrow{BC}$ and $\overrightarrow{CD}$:
$$\overrightarrow{BD} = \overrightarrow{BC} + \overrightarrow{CD} = (2\overrightarrow{a} + 8\overrightarrow{b}) + 3(\overrightarrow{a} - \overrightarrow{b}) = 5\overrightarrow{a} + 5\overrightarrow{b} = 5(\overrightarrow{a} + \overrightarrow{b}).$$
This implies that:
$$\overrightarrow{BD} = 5\overrightarrow{AB}.$$
Hence, vector $\overrightarrow{BD}$ is collinear with vector $\overrightarrow{AB}$, and since point B lies on both these vectors, we can conclude that points A, B, and D are collinear.

(2) Since $k\overrightarrow{a} + \overrightarrow{b}$ and $2\overrightarrow{a} + k\overrightarrow{b}$ are collinear, there exists a scalar $\lambda$ such that:
$$k\overrightarrow{a} + \overrightarrow{b} = \lambda(2\overrightarrow{a} + k\overrightarrow{b}).$$
By comparing the coefficients of $\overrightarrow{a}$ and $\overrightarrow{b}$, we can set up the system of equations:
$$\begin{cases} k = 2\lambda, \\ 1 = \lambda k. \end{cases}$$
Solving this system, we multiply the two equations to eliminate $\lambda$:
$$k = 2 \cdot \frac{1}{k} \implies k^2 = 2 \implies k = \pm\sqrt{2}.$$
Therefore, the value of $k$ is $\boxed{\pm\sqrt{2}}.$