Regarding the question about the Steenrod algebra, it is not true that every non-negatively graded module for the $\text{mod }2$ Steenrod algebra is a direct sum of indecomposable modules. I haven't checked the $\text{mod }p$ Steenrod algebra for odd $p$, but I would be astonished if something similar didn't work.

First, note that not every (ungraded) $\mathbb{F}_2[x]$-module is a direct sum of indecomposables. This follows from more general results, since $\mathbb{F}_2[x]$ is not Artinian, or (more directly) at least one of the proofs of the corresponding fact about a countably infinite product of copies of $\mathbb{Z}$ also shows that a countably infinite product of copies of $\mathbb{F}_2[x]$ is not a direct sum of indecomposable $\mathbb{F}_2[x]$-modules.

Next, I will describe a full exact embedding of the category of $\mathbb{F}_2[x]$-modules into the category of non-negatively graded modules for the Steenrod algebra. So applying this to the example above will give a non-negatively graded module for the Steenrod algebra that is not a direct sum of indecomposables.

Let $V$ be a $\mathbb{F}_2[x]$-module. I will contruct a graded module $\bigoplus_{n\geq0}V_n$ for the Steenrod algebra with
$$V_n=\begin{cases}V&(0\leq n\leq3)\\
0&\text{(otherwise).}
\end{cases}$$

The only Steenrod squares $\text{Sq}^i$ that can act nontrivially are for $i\leq3$, so the only Adem relations that will need to be checked are $\text{Sq}^1\text{Sq}^1=0$ and $\text{Sq}^1\text{Sq}^2=\text{Sq}^3$.

When $i<j$ I'll write $S_{i,j}$ for $\text{Sq}^{j-i}$ considered as a map $V_i\to V_j$. Setting
$$S_{0,1}=S_{2,3}=S_{1,3}=\text{id}_V,$$
$$S_{1,2}=0,$$
$$S_{0,3}=S_{0,2}=x,$$
the Adem relations are satisfied, and so we have constructed the required embedding.