Suppose that $T : M_{2×2}(\Bbb R) \rightarrow U_{2×2}(\Bbb R)$ and $S : U_{2×2}(\Bbb R) \rightarrow P_3$ are linear transformations. Then the composition map $S \circ T : M_{2×2}(\Bbb R) \rightarrow P_3$ is never one-to-one.
To prove this is wrong, I have created a transformation $T$ and $S$ such that the standard matrix for the transformation $S \circ T$ has a pivot in every column, and thus, one-to-one. However, I am not sure if I am on the right track or missing something, as this answer seems too easy.
Following your comment clarifying the notation:
$M_{2\times2}(\Bbb{R})$ ($2\times2$ matrices with real number entries) is a $4$-dimensional vector space over $\Bbb{R}$. $U_{2\times2}(\Bbb{R})$ (upper-triangular $2\times2$ matrices with real number entries) is a $3$-dimensional vector space over $\Bbb{R}$.
Any linear transformation $T : M_{2\times2}(\Bbb{R}) \to U_{2\times2}(\Bbb{R})$ must have a kernel of dimension at least $1$. So for any real vector space $V$ and any linear transformatiom $S : U_{2\times2}(\Bbb{R}) \to V$, the composite $S \circ T$ will have a non-trivial kernel and hence will not be one-one.