I've reviewed the following answer Proving transitivity of $\vDash$ and and $\vdash$ and I'm having trouble grasping the concept. I've tried to recreate the Proof below:
Let $\Gamma \vdash B_1, \Gamma \vdash B_2, ..., \Gamma \vdash B_n$
Let $B_1, B_2, ..., B_n \vdash A$
Prove: $\Gamma \vdash A$
By assumption we have the following $\Gamma$-proofs:
$$\Gamma \cup \{B_1\}$$ $$\Gamma \cup \{B_2\}$$$$...$$$$\Gamma \cup \{B_n\}$$ therefore we know the $\Gamma$-theorems: $B_1, B_2, ..., B_n \subset\Gamma$ by concatenating the above. With (2) we have the following $\Gamma$-proof:$$B_1, B_2, ..., B_n \cup \{A\}$$
Because $B_1, B_2, ..., B_n \subset \Gamma$ and $B_1, B_2, ..., B_n$ are written in $\{A\}$, $B_1, B_2, ..., B_n$ are legitimate in the $\Gamma$-proof context, so $A$ is a $\Gamma$-theorem.
I'd put it differently: you have $n$ proofs that each only use axioms from $\Gamma$ and have $B_i$ as their respective conclusions.
You start the new proof by copying all these old proofs (we still have a finite proof). Now we copy the proof that uses only $B_1, B_2, \ldots, B_n$ as assumptions and $A$ as its final conclusion, and we can refer back to the $B_i$ in the first part of the proof in this final part, instead of using them as assumptions. So the final proof only has assumptions from $\Gamma$ (in the first part) and $B$ as its conclusion, using only valid steps. So $\Gamma \vdash A$.
I don't know this $\Gamma \cup \{A\}$-proof concept, it's not (to me) standard. BTW it's not always true that $B_1, B_2, \ldots, B_n \subset \Gamma$ (which is a nonsense notation anyway: a sequence a subset of a set?).