This question was restated as follows:
Let $V\colon[a,b]\to\mathbb{R}$ be smooth, strictly decreasing and $V(b) = 0$. Suppose that $f\colon[a,b]\to\mathbb{R}$ is smooth and satisfies $f''(x)+V(x) f(x)=0$ on $[a,b]$. Can we then bound $\sup_{x \in [a,b]} |f(x)|$ in terms of $f(a),f'(a),f(b),V(a)$?
This question was also answered there (negatively).
In a comment to that answer, user username asked: "And if $b-a$ is fixed?"
Below is a (positive) answer to this latter question.
We suppose that $a<b$. Without loss of generality (wlog) $f(a)\ge0$. If $f(a)=0$ and $f'(a)=0$, then $f=0$ (on $[a,b]$), so that this case is trivial. So, wlog either $f(a)>0$ or $f(a)=0<f'(a)$. It follows that $f>0$ in a right neighborhood of $a$.
If $f\ge0$ on the entire interval $[a,b]$, then $f''=-Vf\le0$ on $[a,b]$, so that $f$ is concave on $[a,b]$ and hence $0\le f(x)\le f(a)+f'(a)(x-a)\le f(a)+f'(a)_+(b-a)$ for $x\in[a,b]$ (where $u_+:=\max(0,u)$), so that $f$ is bounded on $[a,b]$ in terms of $f(a),f'(a),b-a$.
So, wlog $f$ has a root in $(a,b]$. Let $a_1$ be the smallest of such roots; such a root exists, since $f>0$ in a right neighborhood of $a$. Reasoning as in the previous paragraph, we see
\begin{equation}
0\le f\le m_f:=f(a)+f'(a)_+(b-a)
\end{equation}
on $[a,a_1]$, so that $f$ is bounded on $[a,a_1]$ in terms of $f(a),f'(a),b-a$.
Also, \begin{equation} 0\ge f'(a_1)=f'(a)+\int_a^{a_1}f''(x)\,dx=f'(a)-\int_a^{a_1}V(x)f(x)\,dx \ge f'(a)-\int_a^{a_1}V(a)m_f\,dx\ge f'(a)-(b-a)V(a)m_f. \end{equation} So, $f'(a_1)$ is bounded in terms of $f(a),f'(a),b-a,V(a)$.
If $a_1$ is the only root of $f$ in $(a,b]$, then $f$ is convex on $[a_1,b]$ and hence $0\ge f\ge(b-a_1)f'(a_1)\ge(b-a)f'(a_1)$ on $[a_1,b]$, so that $f$ is bounded on $[a_1,b]$ in terms of $f(a),f'(a),b-a,V(a)$, which implies that $f$ is bounded on $[a,b]$ in terms of $f(a),f'(a),b-a,V(a)$.
To deal with the case when $a_1$ is not the only root of $f$ in $(a,b]$, we need
Lemma 1: Let $V\colon[a,\infty)\to\mathbb{R}$ be nonnegative and nonincreasing. Suppose that $f\colon[a,\infty)\to\mathbb{R}$ is such that $f''+Vf=0$ on $[a,\infty)$. Suppose that $a\le x_1<x_2<\infty$ and $f(x_1)=f(x_2)=0<f(x)$ for $x\in(x_1,x_2)$. Then $|f'|\le|f'(x_1)|$ on $[x_1,x_2]$.
By Lemma 1, $|f'|\le|f'(a_1)|$ on $[a_1,b]$, and hence $|f|\le|f'(a_1)|(b-a_1)\le|f'(a_1)|(b-a)$ on $[a_1,b]$. Thus, again $f$ is bounded on $[a,b]$ in terms of $f(a),f'(a),b-a,V(a)$.
It remains to give
Proof of Lemma 1: The conditions $f(x_1)=f(x_2)=0<f$ on $(x_1,x_2)$ and $f''+Vf=0$ on $[a,\infty)$, together with nonnegativity of $V$, imply that $f$ is concave on $[x_1,x_2]$. So, it attains a maximum $m$ at some point $c\in(x_1,x_2)$. Wlog, $m>0$. So, $f$ is strictly and continuously increasing on $I_1:=[x_1,c]$, and $f$ strictly and continuously decreasing on $[c,x_2]$. So, for each $j\in\{1,2\}$, there is a homeomorphism $p_j\colon[0,m]\to I_j$ such that $p_j(f(x))=f'(x)$ and hence $f''(x)=p_j'(f(x))p_j(f(x))$ for all $x\in I_j$. Thus, for each $j\in\{1,2\}$, we have the ODE $p_j'(y)p_j(y)+v_j(y)y=0$ for $y\in[0,m]$, where $v_j:=V\circ (f_j^{-1})$ and $f_j:=f|_{I_j}$, the restriction of $f$ to $I_j$. Noting that $p_j'p_j=(p_j^2)'/2$ and then integrating the ODE $p_j'(y)p_j(y)+v_j(y)y=0$, for all $y\in[0,m]$ we get \begin{equation} p_j(y)^2=2\int_y^m v_j(z)z\,dz, \end{equation} since $p_j(m)=f'(c)=0$. The condition that $V$ is nonincreasing implies $v_1\ge v_2$. So, $p_1^2\ge p_2^2$ on $[0,m]$, whence $\max_{x\in I_2}|f'(x)|\le\max_{x\in I_1}|f'(x)|$. But, by the concavity of $f$, $\max_{x\in I_1}|f'(x)|=\max_{x\in I_1}f'(x)=f'(x_1)=|f'(x_1)|$. This completes the proof of Lemma 1.
Thus, $f$ is bounded on $[a,b]$ in terms of $f(a),f'(a),b-a,V(a)$. (To bound $f$, we do not need the value of $f(b)$. We also do not need $V$ to be smooth or vanishing at $b$, and we do not need the decrease of $V$ to be strict.)