A first course in noncommutative rings by Lam - Proposition 1.17 about triangular rings. Let $R,S$ be two rings and $M$ be a $(R,S)$-bimodule (i.e. $M$ is a left $R$-module and a right $S$-module such that $(rm)s = r(ms)$ for all $r\in R, s \in S, m \in M$). We can define the ring
$$A= \begin{pmatrix}R \ M\\ 0 \ S\end{pmatrix}$$
with componentwise addition and multiplication of matrices as you would expect it (formal matrix multiplication).
I checked that $A$ becomes an associative ring (with identity if $R$ and $S$ have identity) for these operations.
I have a couple of questions:
(1) Lam says 
"First, it is convenient to identify $R,S$ and $M$ as subgroups in $A$ (in the obvious way) and to think of $A$ as $R \oplus M \oplus S$."

Does this mean that we identity for example $S$ with the subgroup
   $\begin{pmatrix}0 \ 0\\ 0 \ S\end{pmatrix}$ of $A$?  Then $A= R \oplus
 M \oplus S$ indeed becomes true.

(2) Proposition $(1.17)$ states the following:
"The left ideals of $A$ are of the form $I_1 \oplus I_2$ where $I_2$ is a left ideal in $S$ and $I_1$ is a left $R$- submodule of $R \oplus M$ containing $MI_2$."

What does the notation $MI_2$ mean?

 A: Just as with ideal products, this was not intended to mean simply the set of elementwise products $\{xy\mid x\in M, y\in I_2\}$.  
Surely since $I_2\subset S$ and $M$ is a right $S$ module, $MI_2\subseteq M$, but we are talking about left submodules of $M$, so we should expect something that is definitely at least a left $R$ submodule.
Again, using ideal products as our model, $MI_2:=\{\sum m_ia_i\mid m_i\in M, a_i\in I_2\}$, that is, the finite sums of pairwise products, can easily be seen to be a left $R$ submodule of $M$.  (This amounts to the same thing mentioned in the comments, the $\mathbb Z$-span of the pairwise products.)
So when I think about this condition "$I_1$ is a left $R$ submodule of $R\oplus M$ containing $MI_2$" I usually think of it as $$\{0\}\oplus MI_2\subseteq I_1\subseteq R\oplus M$$

And when they say that $I_1$ s a submodule of $⊕$. Is this module structure on $⊕$ usual scalar multiplication on matrices?

No: you would have a problem doing that since for some $\alpha, r\in R$ and $m\in M$, $s\in S$, $\alpha\begin{bmatrix}r&m\\0&s\end{bmatrix}=\begin{bmatrix}\alpha r&\alpha m\\0&\alpha s\end{bmatrix}$ would necessitate an action of $R$ on the left of $S$, something which you don't have necessarily.
It is enough to think of $R\oplus M$ as the regular direct sum of left $R$ modules, so that $\alpha(r,m)=(\alpha r, \alpha m)$. It is not safe to think of the left action going all the way across to the third factor $S$, as I mentioned, but you can model it with matrices in the following way.
The action we just described can be reflected in matrix multiplication if $\alpha\begin{bmatrix}r & m \\ 0 & s\end{bmatrix}:=\begin{bmatrix}\alpha&0\\ 0&1\end{bmatrix}\begin{bmatrix}r&m\\0&s\end{bmatrix}$ and likewise on the right hand side, if $\beta\in S$ then $\begin{bmatrix}r & m \\ 0 & s\end{bmatrix}\beta :=\begin{bmatrix}r&m\\0&s\end{bmatrix}\begin{bmatrix}1&0\\ 0&\beta\end{bmatrix}$ for the natural right action of $S$ on $M\oplus S$.
