Let $R$ be a ring with unity (not neccessarily commutative), and say we have a finite direct sum $$\bigoplus_{i = 1}^{n}R$$ and a submodule $$K \subseteq \bigoplus_{i = 1}^{n} R.$$ If we now view $R$ with the natural $R$-module structure, we know that the direct sum can also be given an $R$-module structure.
Is it the case the projection $$\pi_i:K \to K_j$$ is a submodule of $R$, and hence an ideal of $R$ (with $R$ having the $R$-module structure)?
We have that $$0 \in K_j \implies K_j \neq \emptyset$$ and furthermore, if $$x,y \in K_j, r \in R \implies x+ry \in K_j$$ where we identify an element $k_j$ as being in $K_j$ if there exist an element $k \in K$ such that it's j:th component is $k_j$.
I am not entirely sure about this, it feels like something is wrong with my definition, but I can´t quite put my finger on it.
Note that I only handle the finite case here, but I don´t think the infinite case necessarily changes things.
Yes, this is fine. The following more general facts are true:
If $f : M \to N$ is an $R$-module homomorphism, then the image $f(K)$ of a submodule $K \subseteq M$ is a submodule of $N$.
If $\bigoplus_{i \in I} M_i$ is a direct sum (possibly infinite) of $R$-modules, then for each $j$ the projection $\pi_j : \bigoplus_i M_i \to M_j$ to $M_j$ is an $R$-module homomorphism. (The same is true for products $\prod_{i \in I} M_i$, again possibly infinite.)
A left resp. right ideal of $R$ is a left resp. right submodule of $R$.
But! Note that $K$ is not the direct sum of the $K_j$.