[proofplan]
We use the stated truncated Vaughan convolution identity pointwise and multiply it by an arbitrary arithmetic weight $f(n)$. Summing over $n\le x$ gives four finite sums. The first three have a bounded divisor variable $d\le U$ or $m\le V$, so they are Type I sums. The fourth is bilinear in $m$ and $r$, with $m>V$ and $mr\le x$, so it is the Type II term described in the statement.
[/proofplan]
[step:Record the pointwise identity]
By the hypothesis, for every $n\ge1$,
\begin{align*}
\Lambda(n)=\sum_{\substack{d\mid n,\ d\le U}}\mu(d)\log\frac{n}{d}-\sum_{\substack{dm\mid n,\ d\le U,\ m\le V}}\mu(d)\Lambda(m)+\sum_{\substack{m\mid n,\ m\le V}}\Lambda(m)+\sum_{\substack{mr=n,\ m>V}}\Lambda(m)\left(\varepsilon(r)-\sum_{\substack{d\mid r,\ d\le U}}\mu(d)\right).
\end{align*}
[guided]
This step records the assumed identity in the exact form used later. For each $n\ge1$, the von Mangoldt value $\Lambda(n)$ is written as four finite pieces: the divisor sum with $d\mid n$ and $d\le U$, the correction sum with $dm\mid n$, $d\le U$, and $m\le V$, the short divisor sum with $m\mid n$ and $m\le V$, and the bilinear sum over $mr=n$ with $m>V$. The fourth coefficient is $\varepsilon(r)-\sum_{\substack{d\mid r,\ d\le U}}\mu(d)$.
[/guided]
[/step]
[step:Multiply by an arithmetic weight and sum]
Let $f:\mathbb N\to\mathbb C$ be any arithmetic function and let $x\ge1$. Multiplying the pointwise identity by $f(n)$ and summing over integers $n\le x$ gives
\begin{align*}
\sum_{n\le x}\Lambda(n)f(n)=S_1-S_2+S_3+S_4,
\end{align*}
where
\begin{align*}
S_1:=\sum_{n\le x}f(n)\sum_{\substack{d\mid n,\ d\le U}}\mu(d)\log\frac{n}{d},
\end{align*}
\begin{align*}
S_2:=\sum_{n\le x}f(n)\sum_{\substack{dm\mid n,\ d\le U,\ m\le V}}\mu(d)\Lambda(m),
\end{align*}
\begin{align*}
S_3:=\sum_{n\le x}f(n)\sum_{\substack{m\mid n,\ m\le V}}\Lambda(m),
\end{align*}
and
\begin{align*}
S_4:=\sum_{\substack{mr\le x,\ m>V}}\Lambda(m)\left(\varepsilon(r)-\sum_{\substack{d\mid r,\ d\le U}}\mu(d)\right)f(mr).
\end{align*}
[guided]
After multiplying by $f(n)$ and summing over $n\le x$, the left side becomes $\sum_{n\le x}\Lambda(n)f(n)$. The first three right-hand terms are
\begin{align*}
S_1:=\sum_{n\le x}f(n)\sum_{\substack{d\mid n,\ d\le U}}\mu(d)\log\frac{n}{d},
\end{align*}
\begin{align*}
S_2:=\sum_{n\le x}f(n)\sum_{\substack{dm\mid n,\ d\le U,\ m\le V}}\mu(d)\Lambda(m),
\end{align*}
and
\begin{align*}
S_3:=\sum_{n\le x}f(n)\sum_{\substack{m\mid n,\ m\le V}}\Lambda(m).
\end{align*}
The fourth term is rewritten by putting $n=mr$, giving
\begin{align*}
S_4:=\sum_{\substack{mr\le x,\ m>V}}\Lambda(m)\left(\varepsilon(r)-\sum_{\substack{d\mid r,\ d\le U}}\mu(d)\right)f(mr).
\end{align*}
Thus $\sum_{n\le x}\Lambda(n)f(n)=S_1-S_2+S_3+S_4$.
[/guided]
[/step]
[step:Identify the Type I and Type II structures]
The sum $S_1$ has a divisor variable $d$ constrained by $d\le U$. The sums $S_2$ and $S_3$ have a divisor variable $m$ constrained by $m\le V$. These are Type I expressions in the usual sense: after fixing the bounded variable, the remaining summation is over multiples of that variable.
The sum $S_4$ is bilinear in the factorization variables $m$ and $r$, subject to $m>V$ and $mr\le x$, with coefficient
\begin{align*}
\varepsilon(r)-\sum_{\substack{d\mid r,\ d\le U}}\mu(d).
\end{align*}
This is the asserted Type II expression. Hence the weighted sum $\sum_{n\le x}\Lambda(n)f(n)$ has exactly the decomposition claimed in the statement.
[guided]
The Type I pieces are $S_1$, $S_2$, and $S_3$. In $S_1$ the bounded divisor is $d\le U$. In $S_2$ the bounded variables satisfy $d\le U$ and $m\le V$, and in $S_3$ the bounded divisor is $m\le V$. The Type II piece is
\begin{align*}
S_4=\sum_{\substack{mr\le x,\ m>V}}\Lambda(m)\left(\varepsilon(r)-\sum_{\substack{d\mid r,\ d\le U}}\mu(d)\right)f(mr).
\end{align*}
It is bilinear in $m$ and $r$, has the range $m>V$ and $mr\le x$, and has coefficient $\varepsilon(r)-\sum_{\substack{d\mid r,\ d\le U}}\mu(d)$ depending on $r$. Therefore the decomposition has exactly the Type I and Type II structure claimed.
[/guided]
[/step]
