Sometime one has to left-multiply, sometimes one has to right-multiply. This really depends.
Prerequisites:
You are performing scaling, rotation, and translation. So let us assume we have linear point transformation of the general form:
$$\mathtt T = \left[ \begin{array}{cc} s\mathtt R & \mathbf t \\ \mathtt O& 1\end{array} \right]$$
which first rotates a point by $\mathtt R$ , then scales it by $s$ and then adds the translation $\mathbf t$:
$$\mathtt T \cdot \left( \begin{array}{c} \mathbf x \\ 1\end{array} \right) = \left[ \begin{array}{c} s(\mathtt R\cdot \mathbf x) + \mathbf t \\ 1\end{array} \right]$$
(Note that rotation and scaling commutes: $s(\mathtt R\cdot \mathbf x)=\mathtt R(s\cdot \mathbf x)$)
From now on we will assume that all points $\mathbf y$ are homogenous points ($\mathbf y= (\mathbf x, 1)^\top $).
Mind the reference frames: In order to make it clear whether you need a left or right multiplication, it is important to highlight in which reference frame your points are!
Let us assume, we have points $\mathbf y_a$ in reference frame $a$, and you want to transform them into reference frame $b$, you do
$$ \mathbf y_b = \mathtt T_{ba} \mathbf y_a$$
where $\mathtt T_{ba}$ is a transformation to $b$ from $a$. Note that the indices must match!
Now, let us look at a more complicated example. One might be interested in:
$$\mathbf y_a = \mathtt T_{ab}\mathtt T_{bc}\mathtt T_{cd}\mathbf y_d$$
Further, let's assume that we receive the poses in order (First $\mathtt T_{ab}$, then $\mathtt T_{bc}$...).
We would calculate in an algorithm:
$\mathtt T_{ai} := \mathtt T_{ab}$
(thus, $i=b$)
$\mathtt T_{ai} := \mathtt T_{ai}\cdot \mathtt T_{bc}$
(now, $i=c$)
$\mathtt T_{ai} := \mathtt T_{ai}\cdot \mathtt T_{cd}$
($i=d$)
Thus, we right-multiplied and $\mathtt T_{ai}$ means now $\mathtt T_{ad}$, the transformation from $d$ to $a$. Finally, we can transform our points:
$$\mathbf y_a := \mathtt T_{ad} \mathbf y_d $$
However, if one really wants to left-multiply, this is possible too! Note that $\mathtt T_{ia}=\mathtt T_{ai}^{-1}.$ Thus, we can do:
$\mathtt T_{ia} := \mathtt T_{ab}^{-1}$
($i=b$)
$\mathtt T_{ia} := \mathtt T_{bc}^{-1} \mathtt T_{ia}$
($i=c$)
$\mathtt T_{ia} := \mathtt T_{cd}^{-1}\mathtt T_{ia}$
($i=d$)
Thus, we have $\mathtt T_{ia} = \mathtt T_{da}$, and therefore we can transfrom the point from $d$ to $a$ using the inverse:
$$\mathbf y_a := \mathtt T_{da}^{-1} \mathbf y_d $$
I think there should be a $x$ in front of $b$ in your first equation defining the action of $T$.
To get the matrix form of $T$ with respect to the bases $B$ and $B'$, we must evaluate $T$ on each element of $B$ and write the result in term of elements of $B'$.
To be more precise, let $\mathrm{e}_1:=1-x+x^2$, $\mathrm{e}_2:=1-x$ and $\mathrm{e}_3:=-1$ ; we compute :
$$T(1)=x^2=\mathrm{e}_1-\mathrm{e}_2,$$
$$T(x)=x=-\mathrm{e}_2-\mathrm{e}_3,$$
$$T(x^2)=T(x^3)=0.$$
Thus the representative matrix of $T$ relatively to the bases $B$ and $B'$ is given by
\begin{align}
\begin{pmatrix}1&0&0&0
\\-1&-1&0&0\\
0&-1&0&0
\end{pmatrix}.
\end{align}
We can see that the first column is the coordinates vector of $T(1)$ in the basis $B'$, that the second column is the coordinates vector of $T(x)$ in this same basis, and so on.
Best Answer
$S\circ T$ could be the identity. Let $S:\mathbb R^3\to \mathbb R^2$ be given by $S(x,y,z)=(x,y)$. And let $T:\mathbb R^2\to\mathbb R^3$ be given by $T(x,y)=(x,y,0)$. Then $S\circ T=I_{\mathbb R^2}$.
On the other hand, we can't get $T\circ S$ to be the identity of $\mathbb R^3$, because the rank-nullity theorem tells us that the rank of $T:\mathbb R^2\to \mathbb R^3$ is at most $2=\operatorname{dim}\mathbb R^2$. But the dimension of the image equals the rank. This means $T$ can't be surjective. Hence neither can $T\circ S$.