Having establish the the quadrature formula by cubic interpolation of $f$ based at nodes $(a,f(a))$, $(a+h,f(a+h))$, $(a+2h,f(a+2+2h))$ and $(b,f(b))$ with $h=\frac{b-a}{3}$ as well as its write this quadrature formula in the form
$$\int^b_af\approx\frac{3h}{8}\Big(f(a)+3f(a+h)+3f(a+2h)+f(b)\Big)$$
with error $e_h=-\frac{(b-a)^5}{6480}f^{(iv)}(c)$ for some $c\in (a, b)$.
Divide the interval $[a,b]$ in $3m$ pieces of same length $h=\frac{b-a}{3m}$. This defines a partition $x_k=a+\frac{b-a}{3m}$.
On the each subinterval of the form $[x_{3k},x_{3(k+1)}]$, $k=0,\ldots ,m-1$ apply Simpson's rules based on cubic interpolation (the fist formula in the OP). That is
$$\begin{align}
\int^b_af=\sum^{m-1}_{k=0}\int^{x_{3(k+1)}}_{x_{3k}}f\tag{1}\label{one}
\end{align}$$
and
$$I_k:=\int^{x_{3(k+1)}}_{x_{3k}}f\approx\frac{3h}{8}\Big(f(x_{3k})+3f(x_{3k+1})+3f(x_{3k+2})+f(x_{3(k+1)})\Big)=A_k$$
Denoting the righthand side of \eqref{one} by $A$ we have that
\begin{align}
\frac{8}{3h}A&=\sum^{m-1}_{k=0}\Big(f(x_{3k})+3f(x_{3k+1})+3f(x_{3k+2})+f(x_{3(k+1)})\Big)\\
&=f(x_0)+\sum^{m-1}_{k=0}3\big(f(x_{3k+1})+f(x_{3k+2})\big) +2\sum^{m-1}_{k=1}f(x_{3k}) + f(x_{3m})
\end{align}
The composite formula follows.
As for the error, each approximation $A_k$ to $I_k$ has error $-\frac{3}{80} h^5 f^{(iv)}(c_k)$ for some $c_k\in (x_{3k},x_{3(k+1)})$. The sum of all errors is
$$E:=-\frac{3}{80}h^5\sum^{m-1}_{j=0}f^{(iv)}(c_k)$$
If we assume that $f\in\mathcal{C}^4([a,b])$, there there is $c\in(a, b)$ such that $f^{(iv)}(c)=\frac1m\sum^{m-1}_{k=0}f^{(iv)}(c_k)$ and so
$$
E=-\frac{1}{80}(b-a)h^4f^{(iv)}(c)
$$
With regards to the error in this quadrature, there are somewhat explicit formulas for the errors resulting from interpolation methods on evenly distributed nodes:
For $n\in\mathbb{N}$, $n\geq2$, the error of approximation to $\int^b_af$ in the quadrature method based on interpolating over the uniform partition $a_k=\frac{k}{n}(b-a)$, $0\leq k\leq n$, is
- If $n$ is even and $f\in C^{(n+2)}[a,b]$,
$$E_n=\frac{K_n}{(n+2)!}f^{(n+2)}(\eta),\quad K_n=\int^b_a x\prod^n_{k=0}(x-x_k)\,dx$$
for some $a<\eta<b$,
- If $n$ is odd and $f\in C^{(n+1)}[a,b]$,
$$E_n=\frac{K_n}{(n+1)!}f^{(n+1)}(\eta),\qquad K_n=\int^b_a\prod^n_{j=1}(x-x_j)\,dx$$
For Simpson's 3/8 rule, $n=3$ and so
$K_3=\int^b_a (x-a)(x-\tfrac{2a+
b}{3})(x-\tfrac{a+2b}{3})(x-b)\,dx$
The simplification, though messy with pencil and paper, yields
$E_3=-\frac{(b-a)^5}{6480}f^{(4)}(\eta)$ for some $a<\eta<b$.
Reference: Krylov, V. I. (translated by Stroud, A.), Approximate calculation of integrals, Dover Publications. 2005 pp 89-91
