Left Homotopy is an Equivalence Relation I was reading Mark Hovey's Model Categories and I am confused about the following proof for left homotopy being an equivalence relation

Firstly, how does $B$ being cofibrant imply that $t$ is a weak equivalence? I was thinking maybe we could show it is the retract of a weak equivalence, or use the "2-of-3 axiom".
Secondly, are the maps $j_0$ and $j_1$ defined correctly, or are the indicies on the $i$'s switched?
 A: 
Firstly, how does $$ being cofibrant imply that $t$ is a weak equivalence?

Since $s$ is always a weak equivalence, the only interesting statement here is that the map $B^\prime \to C$ (and, by the same argument, also $B^{\prime\prime} \to C$) is a trivial cofibration.  This is because the morphism $t$, which is induced by the universal property for $C$, ensures that the diagram
$$\require{AMScd}
\begin{CD}
B^\prime @>>> C \\
@VsVV @VtVV\\
B @>1_B>> B
\end{CD}$$
commutes so that $t$ is a weak equivalence by 2-out-of-3.
For the interesting statement, we observe that because $B$ is cofibrant and cofibrations are stable under pushout, both the coproduct inclusions
$$\require{AMScd}
\begin{CD}
    0 @>>> B\\
    @VVV @VVV\\
    B @>>> B\coprod B
\end{CD}$$
are cofibrations.  The morphism $i_0^\prime \colon B \to B^{\prime\prime}$ is the composition of the cofibration $B \coprod B \to B^{\prime\prime}$ with one of these inclusions, hence is itself a cofibration.  Moreover, we observe that $i_0^\prime$ is a weak equivalence because the morphism $s^\prime \colon B^{\prime\prime} \to B$ is a weak equivalence and $s \circ i_0^\prime = 1_B.$ 
Now apply stability of trivial cofibrations under pushout to the commutative diagram in question to see that $B^\prime \to C$ is a trivial cofibration. 
