The above is from paper of Selberg on the elementary proof of prime number theorem. Prior to eq.$(2.11)$ the author derives an asymptotic expression for $\sum_{pq≤x}\ln(p)\ln(q)$, where $p,q$ are primes. For that he makes use of eq.$(2.10)$.
In eq.$(2.10)$ the remainder term is $O\left(\frac{x}{\ln(x)}\right)$. So using it for $\sum_{pq≤x}\ln(p)\ln(q)$ should result in the remainder term $O\left(\frac{x/p}{\ln(x/p)}\right)$
My question:-
(1) Why did the author take the remainder term to be $O\left(\frac{x/p}{1+\ln(x/p)}\right)$?
(2) How does that become $O(x\ln\ln(x))$ by using abel summation?
As noted in comments,
$$O\!\left( {\frac{x}{{\ln x}}} \right) = O\!\left( {\frac{x}{{1 + \ln x}}} \right).$$
The reason why we the author adds $1$ is because the term when $p = x$ is in the sum and $\log(x/p)$ would be $0$ — and we don't want to divide by zero.
The second question concerns showing $$ x \sum_{p \leq x} \frac{\log p}{p (1 + \log x/p)} \ll x \log \log x. $$ For this, we convert to a Riemann-Stieltjes integration problem (approximately equivalent to summation by parts or Abel summation). For this it will be convenient to note that the derivative $$ \frac{d}{dt} \frac{\log t}{t (1 + \log x/t)} = -\frac{\log\left(t\right)}{t^{2} {\left(\log\left(\frac{x}{t}\right) + 1\right)}} + \frac{1}{t^{2} {\left(\log\left(\frac{x}{t}\right) + 1\right)}} + \frac{\log\left(t\right)}{t^{2} {\left(\log\left(\frac{x}{t}\right) + 1\right)}^{2}} \approx -\frac{\log\left(t\right)}{t^{2} {\left(\log\left(\frac{x}{t}\right) + 1\right)}}. $$ Then we have by integration by parts $$ \sum_{p \leq x} \frac{\log p}{p (1 + \log x/p)} = \int_{2}^x \frac{\log t}{t (1 + \log x/t)} d \lfloor \pi(t) \rfloor \lesssim \frac{\log x}{x} \pi(x) - \frac{\log 2}{2} + \int_2^x \pi(t) \frac{\log\left(t\right)}{t^{2} {\left(\log\left(\frac{x}{t}\right) + 1\right)}} dt. $$ This would be a pure equality if I used the whole derivative above, but I extract the dominant contribution here and write $\lesssim$ instead. (Note this includes the only positive contribution). The first term
By the prime number theorem, $\pi(t) \sim t\log t$. Thus \begin{align} \int_2^x \pi(t) \frac{\log\left(t\right)}{t^{2} {\left(\log\left(\frac{x}{t}\right) + 1\right)}} dt &\ll \int_2^x \frac{t}{\log t} \frac{\log\left(t\right)}{t^{2} {\left(\log\left(\frac{x}{t}\right) + 1\right)}} dt = \int_2^x \frac{1}{t {\left(\log\left(\frac{x}{t}\right) + 1\right)}} dt \\ &\ll \int_2^{x-3} \frac{1}{t {\left(\log\left(\frac{x}{t}\right)\right)}} dt = - \log \log (x/t) \Big|_{2}^{x-3} \ll \log \log (x/2) \ll \log \log x. \end{align} I use the final bound $x-3$ above for a similar reason that Selberg uses $1 + \ln x$ — it makes the domain of the log work out (and the final bit of the integral is much smaller than the rest, so we lose nothing).