Proposition Claim: Let $A$ be a finite-dimensional algebra over a field $k$. Suppose T,T' are simple $A$-modules and that there is a non-split extension $M$ of $T$ by T'. Then $T$ and T' are composition factors of the same projective indecomposable module.
Proof. We have a short exact sequence 0\to T'\xrightarrow{\iota_1} M\xrightarrow{\pi_1} T\to 0.
Let $P_T$ be the projective indecomposable corresponding to $T$, so $P_T/\mathrm{rad}(P_T)\cong T$, i.e. we have another short exact sequence $ 0\to \mathrm{rad}(P_T)\xrightarrow{\iota_2}P_T\xrightarrow{\pi_2}T\to 0. $
By the universal property of projective modules, using the surjective maps $\pi_1$ and $\pi_2$, there is a unique map $\varphi:P_T\to M$ satisfying $\pi_1\circ \varphi=\pi_2$.
Since the second sequence is exact, we know $\pi_2\circ\iota_2=0$ and so $\pi_2\circ\varphi\circ\iota_1=0$. Notice that \ker\pi_2=T'.
Hence by the universal property of $\ker\pi_2$, there exists a unique nonzero map \psi:\mathrm{rad}(P_T)\to T'. Since T' is simple, $\psi$ is surjective and by the first isomorphism theorem \mathrm{rad}(P_T)/\ker\psi\cong T'. Hence $P_T$ has a filtration $ P_T\supset \mathrm{rad}(P_T)\supset \ker\psi\supset 0 $ with composition factors $T$ and T' as required.