In the work leading to question 1, you have a mistake. When you count the number of pairs $(p_1,p_2)$ of primes with $p_1\cdot p_2 \leqslant x$ using the hyperbola method, you count a) the pairs where $p_1 \leqslant \sqrt{x}$ (and $p_2 \leqslant x/p_1$), b) the pairs where $p_2 \leqslant \sqrt{x}$ (and $p_1 \leqslant x/p_2$), and then c) you subtract the number of doubly counted pairs, that is the number of pairs with $p_1 \leqslant \sqrt{x}$ and $p_2 \leqslant \sqrt{x}$, not the number of pairs where $p_1\cdot p_2 \leqslant \sqrt{x}$. Thus
$$\sum_{p_1p_2 \leqslant x} 1 = \sum_{p_1 \leqslant \sqrt{x}} \pi\biggl(\frac{x}{p_1}\biggr) + \sum_{p_2 \leqslant \sqrt{x}} \pi\biggl(\frac{x}{p_2}\biggr) - \pi(\sqrt{x})^2.$$
Now $\pi(\sqrt{x})^2$ is easily handled with the prime number theorem, since $\log \sqrt{x} = \frac{1}{2} \log x$, we have
$$\pi(\sqrt{x})^2 \sim \frac{4x}{(\log x)^2}.$$
The weaker but more than sufficient $\pi(\sqrt{x})^2 \in \mathcal{O}\bigl(\frac{x}{(\log x)^2}\bigr)$ follows from Chebyshev's bounds.
Since the number of pairs with $p_1 \cdot p_2 \leqslant \sqrt{x}$ is smaller than $\pi(\sqrt{x})^2$ (except for very small $x$), the error estimate of $\mathcal{O}\bigl(\frac{x}{\log x}\bigr)$ is also easily established for your sum:
$$\sum_{p \leqslant \sqrt{x}} \pi\biggl(\frac{\sqrt{x}}{p}\biggr) \leqslant \sum_{p \leqslant \sqrt{x}} \pi(\sqrt{x}) = \pi(\sqrt{x})^2.$$
However, once you replace $\pi(\sqrt{x}/p)$ with $\mathcal{O}\bigl(\frac{\sqrt{x}}{p\log (\sqrt{x}/p)}\bigr)$ in that sum, you cannot reach the desired bound any more, since the last term in the sum may blow up. If $x = q^2$ is the square of a prime, the last term is $\mathcal{O}\bigl(\frac{1}{0}\bigr)$, and that doesn't even make sense. If $x$ is not quite the square of a prime, but very close, the last term can still become arbitrarily large.
Thus, if we want to estimate a sum of terms with a factor $\log (u/p)$ in the denominator, we must make sure that $p$ stays significantly smaller than $u$ (how much smaller that must be depends on the situation).
With the above estimate, we have established that
$$\pi_2(x) = \sum_{p \leqslant \sqrt{x}} \pi\biggl(\frac{x}{p}\biggr) + \mathcal{O}\biggl(\frac{x}{\log x}\biggr).\tag{1}$$
We want to show that
$$\pi_2(x) = x\sum_{p\leqslant \sqrt{x}} \frac{1}{p\log \frac{x}{p}} + o\biggl(\frac{x\log \log x}{\log x}\biggr)\tag{2}$$
using the prime number theorem. Clearly the error term in $(1)$ belongs to $o\bigl(\frac{x\log \log x}{\log x}\bigr)$, so we need not care about that and can concentrate on
$$\sum_{p \leqslant \sqrt{x}} \pi\biggl(\frac{x}{p}\biggr).$$
You have done that essentially correctly.
There is a small mathematical problem, you say that for $\varepsilon > 0$ there is an $M$ such that [estimate] for $x > M$. But the argument of $\pi$ is $x/p$, so the prime number theorem just says that [estimate] holds for $x/p > K$. Now in our sum we have $p \leqslant \sqrt{x}$, so $x > K^2$ implies $x/p > K$, and your assertion is true, but the constraint on $p$ that makes it so should be mentioned.
Then there is a non-mathematical problem. You keep too much of Mertens' theorem, so that you don't see that you have already achieved your goal. Just using that (for $y > y_0$) there is a constant $C$ such that $\sum_{p \leqslant y} \frac{1}{p} \leqslant C\log \log y$, and $\log \log \sqrt{x} \leqslant \log \log x$, you get
$$\sum_{p \leqslant \sqrt{x}} \lvert R(x,p)\rvert \leqslant 2C\varepsilon \cdot \frac{x\log \log x}{\log x}\tag{$\ast$}$$
for $x > M$, which is what you want. If you don't see that yet, note that $(\ast)$ implies
$$\limsup_{x\to \infty} \frac{\log x}{x\log \log x}\sum_{p \leqslant \sqrt{x}} \lvert R(x,p)\rvert \leqslant 2C\varepsilon,$$
and since $\varepsilon > 0$ was arbitrary, this means
$$\lim_{x\to \infty} \frac{\log x}{x\log \log x}\sum_{p \leqslant \sqrt{x}} \lvert R(x,p)\rvert = 0,$$
i.e.
$$\sum_{p \leqslant \sqrt{x}} \lvert R(x,p)\rvert \in o\biggl(\frac{x\log \log x}{\log x}\biggr).$$
