Yoga on coherent flat sheaves $mathcal{F}$ over projective space $mathbb{P}^n$

I’m reading Mumfords’s Lectures on Curves on an Algebraic Surface (jstor-link: https://www.jstor.org/stable/j.ctt1b9x2g3)
and I found in Lecture 7 (RESUME OF THE COHOMOLOGY OF COHERENT SHEAVES ON
$mathbb{P}^n$; p 47) dealing with yoga on coherent
sheaves $F$ over pojective space $mathbb{P}^n$ I found on page 52
a proof I not understand:

Corollary 3: Given a coherent sheaf $mathcal{F}$ on $mathbb{P}^n times S$,
$mathcal{F}$ is flat over $S$
if and only if there exists an $m_0$ such that if $m ge m_0$,
$p_* mathcal{F}(m)$ is
locally free. Hence, in this case, the Hilbert polynomial of on
$mathbb{P}^n _s$ is locally constant.

Proof: If $F$ is flat over $S$, then let $m_0$ be large enough so
that derived image $R^i p_*(mathcal{F}(m))=(0)$, if $i>0, m ge m_0$.
Using Corollary $1$ and $1 frac{1}{2}$ one deduces that
$p_*(mathcal{F}(m)) otimes k(s)$ maps onto $H^0(mathbb{P}^n _s, mathcal{F}_s(m))$
for all $sin S, m ge m_0$. Then by (iii), $p_* mathcal{F}(m)$ is locally free.
As for converse […]

In original:

Problem: The "…Then by (iii), $p_* mathcal{F}(m)$ is locally free…" part I not understand.
(iii) (on page 51) states:

By above we know $p_*(mathcal{F}(m)) otimes k(s) to H^0(mathbb{P}^n _s, mathcal{F}_s(m))$
is surjective, that is we can apply (iii) to $i=1$ and deduce
$R^1p_*(mathcal{F}(m))$ is locally free sheaf. But Mumford claims this for
$p_* mathcal{F}(m)= R^0 p_* mathcal{F}(m)$.

Is this an error in the proof or do I miss something?