Update: it is not possible to reply to this question without additional information.
My comment below: "I have to agree with you that one "cannot derive (2) from (1) alone". Now it seems to me that one must consider how the sequences $v_{h,n}^{\prime }$ and $v_{h,n}$ are constructed. That is described in the first part of the paper, but unfortunately it is not easy for me to summarize it. As I understand Apéry transformed repeatidely a continued fraction whose approximants are $\dfrac{u_{h,n}^{\prime }}{u_{h,n}}$, iterating on $h$. The above sequences are $v_{h,n}^{\prime }=\dfrac{u_{h,n}^{\prime }}{h!n!}, v_{h,n}=\dfrac{u_{h,n}}{h!n!}$."
In Irrationalité de Certaines Constantes, Bull. section des sciences du C.T.H.S., n.º3, p.37-53, Roger Apéry derives rational approximations $\dfrac{v_{h,n}^{\prime }}{v_{h,n}}$ for $\ln (1+t)$, $\zeta (2)$ and $\zeta (3)$, the simplest being the one for the $\ln $. The sequences $v_{h,n}^{\prime }$ and $v_{h,n}$, whose ratio converges to $\ln (2)$, satisfy the recursive relation
$(n+1)v_{h,n+1}-(2h+1)v_{h,n}-nv_{h,n-1}=0.\qquad (1)$
The diagonal sequences $w_{n}^{\prime }=v_{n,n}^{\prime },w_{n}=v_{n,n}$ satisfy
$(n+1)w_{n+1}-3\left( 2n+1\right) w_{n}-nw_{n-1}=0.\qquad (2)$
Remarks:
The initial conditions for $v_{h,n}^{\prime }$, $v_{h,n}$, $w_{n}^{\prime }$, $w_{n}$ are not indicated in the paper.
Recurrences $(1)$ and $(2)$ are the particular case for $t=1$ of, respectively,
$(n+1)v_{h,n+1}-\left( \left( n+1\right) -nt+h\left( 1+t\right) \right) v_{h,n}-ntv_{h,n-1}=0\qquad (\ast)$
(to simplify the notation the index $h$ was deleted in the original) and
$(n+1)w_{n+1}-\left( 2n+1\right) \left( 2+t\right) w_{n}-nt^{2}w_{n-1}=0.\qquad (\ast\ast)$
Question: How do you derive $(2)$ from $(1)$?
Copy of the mentioned paper