I have been asked to show both $M_n(R)$ (the $n\times n$ matrices over a ring $R$) and $T^+_n(R)$ (the $n\times n$ upper triangular matrices over $R$) both decompose into a direct sum of their rows (thought of as right modules). However, this seems quite straightforward, so I am unsure if I am missing something.
This is what I have done so far. Start with $M_n(R)$, and for $1\leq i\leq n$ set $R(i)=\left\{\begin{pmatrix}0&0&\cdots&0\\\vdots&\vdots&&\vdots\\a_{i1}&a_{i2}&\cdots&a_{in}\\\vdots&\vdots&&\vdots\\0&0&\cdots&0\end{pmatrix}\:\:,\:\:a_{ij}\in M_n(R)\right\}$.
Clearly each $R(i)$ is a right $M_n(R)$-module, and $R(i)\cap(R(1)+\cdots R(i-1)+R(i+1)+\cdots+R(n))=\{0\}$ so we can form the direct sum. Further, they both $M_n(R)$ and $\bigoplus R(i)$ seem to be contained in one another. I suppose if we wanted to be really formal we could define the map $\varphi:\bigoplus R(i)\to M_n(R)$ by mapping $\varphi(R_1,\cdots,R_n)=R_1+\cdots+R(n)$, where $R_i\in R(i)$. As we add and scalar multiply component wise, $\varphi$ is clearly an $M_n(R)$-module homomorphism. It is obviously injective and surjective, so it is a bijective. Am I missing anything?
Moreover, how does this change for $T_n^+(R)$? Surely the same argument applies?
Finally, is the above true over rings? And what about infinite matrices? Here, I am assuming only finitely many non-zero entries in an infinite matrix are zero.
What you're doing is correct, and is true over any ring. It is also true for infinite matrices in any context where such matrices make sense.
You shouldn't be surprised that this seems so obvious. Matrix spaces are by definition a free module $R^{n^2}$ with an algebra structure on top, so grouping by rows gets you the direct sum you want on the level of $R$-modules. Then it's basically just the definition of matrix multiplication that the rows are acted on independently from the right, so the $R$-module direct sum respects the algebra structure.
What you need to "show" to get this question correct is now not a question of whether you understand the problem or not, you understand it just fine. It's a question of how much detail to give so that your teacher believes that you understand it, and that depends on your teacher.