This is the info you currently have:
\begin{align*}
\begin{pmatrix} v_1 ' \\ v_2' \end{pmatrix} &= \begin{pmatrix} 0 & 1 \\ -b(x) & a(x) \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}\\[0.3cm]
\begin{pmatrix} w_1 ' \\ w_2' \end{pmatrix} &= \begin{pmatrix} 0 & 1 \\ 0 & -2\frac{v_1'}{v_1} + a(x) \end{pmatrix} \begin{pmatrix} w_1 \\ w_2 \end{pmatrix}
\end{align*}
It can be rewritten as follows:
\begin{align*}
v_1' &= v_2\\
v_2' &= -b(x) v_1 + a(x) v_2 \tag{1}\\
w_1' &= w_2\\
w_2' &= -2\frac{v_1'}{v_1} w_2 + a(x)w_2 \tag{2}
\end{align*}
Equation $(2)$ gives this:
$$ v_1 w_2' = -2 v_1' w_2 + a(x)v_1w_2 $$
And since $w_1' = w_2$, then $w_1'' = w_2'$ and we have:
$$ v_1 w_1'' = -2 v_1' w_1' + a(x)v_1w_1' \tag{3}$$
You want to show this:
$$ (v_1 w_1)'' = -b(x) (v_1w_1) + a(x) (v_1w_1)'$$
Expand on the LHS to get this:
$$ v_1''w_1 + 2v_1'w_1' + v_1w_1'' = -b(x) (v_1w_1) + a(x) (v_1w_1)' \tag{4}$$
What we're going to do is manipulate the LHS of Equation $(4)$ to make it look like the RHS. First substitute Equation $(3)$ into the LHS of Equation $(4)$ above and simplify to get:
$$ v_1''w_1 + a(x) v_1 w_1' = -b(x) (v_1w_1) + a(x) (v_1w_1)' \tag{5}$$
Since $v_1' = v_2$, then $v_1'' = v_2'$ and from Equation $(1)$ above we have:
$$ v_1'' = -b(x) v_1 + a(x) v_1' $$
Now multiply both sides by $w_1$ to get:
$$ v_1''w_1 = -b(x) v_1w_1 + a(x) v_1'w_1 \tag{6} $$
Substitute Equation $(6)$ into the LHS of Equation $(5)$ to get:
$$ -b(x) v_1w_1 + a(x) v_1'w_1 + a(x) v_1 w_1' = -b(x) (v_1w_1) + a(x) (v_1w_1)' \tag{7}$$
Notice we have the $b(x)$ term that we want. Next:
\begin{align*}
a(x)v_1'w_1 + a(x) v_1w_1' &= a(x) \left(v_1'w_1 + v_1w_1'\right)\\
&= a(x) (v_1w_1)'
\end{align*}
The last equality above follows from the product rule. Finally, substitute that into Equation $(7)$ to get the desired result:
$$ -b(x) v_1w_1 + a(x) (v_1w_1)' = -b(x) (v_1w_1) + a(x) (v_1w_1)' $$
