Mertens' third theorem is just the exponentiated version of the second theorem (without the bounds that Mertens proved for his second theorem):
\begin{align}
-\ln\Biggl(\ln n\prod_{p\leqslant n}\biggl(1 - \frac{1}{p}\biggr)\Biggr)
&= -\ln \ln n - \sum_{p\leqslant n} \ln \biggl(1 - \frac{1}{p}\biggr)\\
&= \Biggl(\sum_{p\leqslant n}\frac{1}{p} - \ln \ln n - M\Biggr) + \Biggl(M - \sum_{p\leqslant n} \biggl(\ln\biggl(1-\frac{1}{p}\biggr) + \frac{1}{p}\biggr)\Biggr),
\end{align}
where the first term converges to $0$ by Mertens' second theorem, and the second term converges to $\gamma$ by definition of $M$.
Mertens' bounds in the second theorem and estimates for
$$\sum_{p > n}\biggl(\ln\biggl(1-\frac{1}{p}\biggr)+\frac{1}{p}\biggr)$$
give you bounds for
$$e^\gamma\ln n\prod_{p\leqslant n}\biggl(1-\frac{1}{p}\biggr),\tag{$\ast$}$$
and conversely bounds for that give you bounds for
$$\left\lvert\sum_{p\leqslant n}\frac{1}{p} - \ln \ln n - M\right\rvert,\tag{$\ast\!\ast$}$$
but it is doubtful whether one can directly prove bounds for $(\ast)$ that give you back Mertens' bounds for $(\ast\ast)$.
One can use Mertens' first theorem to derive the second via an integration by parts, Hardy and Wright for example do that, but don't give explicit bounds on $(\ast\ast)$.
For $x > 0$ we define
$$S(x) := \sum_{p\leqslant x} \frac{\ln p}{p}.$$
Mertens' first theorem tells us
$$\lvert S(x) - \ln x\rvert \leqslant 2 + O(x^{-1}),$$
and we can write
$$T(x) := \sum_{p\leqslant x} \frac{1}{p} = \int_{3/2}^x \frac{1}{\ln t}\,dS(t)$$
with a (Riemann/Lebesgue-) Stieltjes integral. Integration by parts yields
\begin{align}
T(x) &= \int_{3/2}^x \frac{1}{\ln t}\,dS(t)\\
&= \frac{S(x)}{\ln x} - \frac{S(3/2)}{\ln \frac{3}{2}} - \int_{3/2}^x S(t)\,d\biggl(\frac{1}{\ln t}\biggr)\\
&= \frac{S(x)}{\ln x} + \int_{3/2}^x \frac{S(t)}{t(\ln t)^2}\,dt\\
&= \frac{S(x)}{\ln x} + \int_{3/2}^x \frac{dt}{t\ln t} + \int_{3/2}^x \frac{S(t) - \ln t}{t(\ln t)^2}\,dt\\
&= \ln \ln x + \underbrace{1 - \ln \ln \frac{3}{2} + \int_{3/2}^\infty \frac{S(t) - \ln t}{t(\ln t)^2}\,dt}_M + \underbrace{\frac{S(x)-\ln x}{\ln x} - \int_x^\infty \frac{S(t)-\ln t}{t(\ln t)^2}\,dt}_{O\bigl(\frac{1}{\ln x}\bigr)}.
\end{align}
I'm not sure, however, whether one can get exactly Mertens' bounds on $(\ast\ast)$ easily from that.
So in a way, Mertens' first theorem is the most powerful, since it implies the others, at least if we don't need explicit bounds for the differences.
Here is a general strategy. You are interested in
$$\sum_{p \leq x} \frac{1}{p^{1 + \frac{2}{\log x}}} - \frac{2}{\log x} \sum_{p \leq x} \frac{\log p}{p^{1 + \frac{2}{\log x}}} + \frac{1}{(\log x)^2} \sum_{p \leq x} \frac{(\log p)^2}{p^{1 + \frac{2}{\log x}}}.$$
We use partial summation, noting that
$$\sum_{p \leq x} \frac{1}{p} = \log \log x + b + O\left(\frac{1}{\log x}\right), \qquad \sum_{p \leq x} \frac{\log p}{p} = \log x + O(1), \qquad \sum_{p \leq x} \frac{(\log p)^2}{p} = \frac{(\log x)^2}{2} + O(1).$$
Here $b$ is some explicit constant.
The first terms via partial summation are
$$e^{-2} \log \log x + b e^{-2}, \qquad -2e^{-2}, \qquad \frac{e^{-2}}{2}.$$
The error is $O(1/\log x)$ for the first two and $O(1/(\log x)^2)$ for the third.
For the second terms, we note that
$$\frac{d}{dt} \frac{1}{t^{\frac{2}{\log x}}} = -\frac{2}{\log x} \frac{1}{t} \exp\left(-\frac{2 \log t}{\log x}\right),$$
and so we must evaluate
$$\frac{2}{\log x} \int_{2}^{x} \log \log t \exp\left(-\frac{2\log t}{\log x}\right) \, \frac{dt}{t}, \qquad \frac{2b}{\log x} \int_{2}^{x} \exp\left(-\frac{2\log t}{\log x}\right) \, \frac{dt}{t}, \qquad -\frac{4}{(\log x)^2} \int_{2}^{x} \log t \exp\left(-\frac{2\log t}{\log x}\right) \, \frac{dt}{t}, \qquad \frac{1}{(\log x)^3} \int_{2}^{x} (\log t)^2 \exp\left(-\frac{2\log t}{\log x}\right) \, \frac{dt}{t}.$$
(There are also error terms; the method below shows that they are $O(\log \log x/\log x)$.) We make the change of variabes $t \mapsto e^t$, so that these become
$$\frac{2}{\log x} \int_{\log 2}^{\log x} \log t \exp\left(-\frac{2 t}{\log x}\right) \, dt, \qquad \frac{2b}{\log x} \int_{\log 2}^{\log x} \exp\left(-\frac{2 t}{\log x}\right) \, dt, \qquad -\frac{4}{(\log x)^2} \int_{\log 2}^{\log x} t \exp\left(-\frac{2 t}{\log x}\right) \, dt, \qquad \frac{1}{(\log x)^3} \int_{\log 2}^{\log x} t^2 \exp\left(-\frac{2 t}{\log x}\right) \, dt.$$
The latter three can be explicitly evaluated (directly for the second, via integration by parts for the last two); they give
$$-b e^{-2} - b \exp\left(-\frac{2 \log 2}{\log x}\right), \qquad 3e^{-2} - \exp\left(-\frac{2 \log 2}{\log x}\right) \left(\frac{2 \log 2}{\log x} + 1\right), \qquad -\frac{5}{4} e^{-2} + \frac{1}{4} \exp\left(-\frac{2 \log 2}{\log x}\right) \left(\frac{2 (\log 2)^2}{(\log x)^2} + \frac{2 \log 2}{\log x} + 1\right).$$
Note that $\exp(-2\log 2/\log x) = 1 + O(1/\log x)$. For the former, we integrate by parts once, getting
$$-e^{-2} \log \log x + \log \log 2 \exp\left(-\frac{2 \log 2}{\log x}\right) + \int_{\log 2}^{\log x} \frac{1}{t} \exp\left(-\frac{2 t}{\log x}\right) \, dt.$$
For this second term, we make the change of variables $t \mapsto \frac{\log x}{2} t$, so that this becomes
$$\int_{\frac{2 \log 2}{\log x}}^{2} \frac{e^{-t}}{t} \, dt = E_1\left(\frac{2 \log 2}{\log x}\right) - E_1(2),$$
where $E_1(z)$ is the exponential integral. Since $E_1(z) = -\log z - \gamma_0 + O(z)$ about $z = 0$, we get additional terms
$$\log \log x - \log 2 - \log \log 2 - \gamma_0 - E_1(2) + O\left(\frac{1}{\log x}\right).$$
So now we combine everything and find that the desired asymptotic is
$$\log \log x - b + \frac{e^{-2}}{4} + \frac{1}{4} - \log 2 - \gamma_0 - E_1(2) + O\left(\frac{\log \log x}{\log x}\right).$$
Alternatively, you can use your approach to get the expression
$$\frac{4}{\log x} \int_{2}^{x} \log \log t \exp\left(-\frac{2 \log t}{\log x}\right) \, \frac{dt}{t} - \frac{6}{(\log x)^2} \int_{2}^{x} \log t \log \log t \exp\left(-\frac{2 \log t}{\log x}\right) \, \frac{dt}{t} + \frac{2}{(\log x)^3} \int_{2}^{x} (\log t)^2 \log \log t \exp\left(-\frac{2 \log t}{\log x}\right) \, \frac{dt}{t}.$$
(Note that you seem to have calculated the derivative incorrectly.) We again make the change of variables $t \mapsto e^t$, yielding
$$\frac{4}{\log x} \int_{\log 2}^{\log x} \log t \exp\left(-\frac{2 t}{\log x}\right) \, dt - \frac{6}{(\log x)^2} \int_{\log 2}^{\log x} t \log t \exp\left(-\frac{2 t}{\log x}\right) \, dt + \frac{2}{(\log x)^3} \int_{\log 2}^{\log x} t^2 \log t \exp\left(-\frac{2 t}{\log x}\right) \, dt.$$
Now integrate by parts repeatedly, integrating the exponential and differentiating everything else.
Best Answer
Let me first lead with that the question, as stated, is nearly a duplicate of this question. However, since this question also asks for more general approaches to problems like this, I'll leave a relatively general approach below that allows one to calculate series like this without too much effort.
Approach
We utilize Riemann-Stieltjes integration. Observe that
$$ \sum_{p\text{ prime}}f(p)=\int_{2^{-}}^\infty f(x)\,d\pi(x) $$ where $\pi(x)=\sum_{p\text{ prime}}1$ is the prime counting function. Thus, if we find a good approximation of $\pi(x)$, we can estimate the value of integrals like this asymptotically using integration by parts. For several problems, the approximation $$\pi(x)=\frac{x}{\log x}+O\left(xe^{-c\sqrt{\log x}}\right)$$ for a positive constant $c$ (better estimates are known but this is sufficient for our purposes). For $p\leq T$, we have \begin{align*} \int_{2^-}^T f(x)\,d\pi(x)&=\int_{2^-}^T f(x)\,d\left(\frac{x}{\log x}+O\left( xe^{-c\sqrt{\log x}}\right)\right)\\ &=\int_{2^-}^Tf(x)\,d\left(\frac{x}{\log x}\right)+\int_{2^{-}}^T f(x)\,d\left(O\left( xe^{-c\sqrt{\log x}}\right)\right)\\ &=\int_2^T f(x)\left(\frac{\log x -1}{\log^2 x}\right)\,dx+\mathcal{E}\tag{1} \end{align*} where $\mathcal{E}$ represents the error. The error can be estimated using integration by parts in the following way: \begin{align*} \int_{2^-}^T f(x)\,dO\left( xe^{-c\sqrt{\log x}}\right)&\ll xf(x)e^{-c\sqrt{\log x}}\bigg]_{x=2}^{x=T}+\int_{2}^T xe^{-c\sqrt{\log x}}\ f'(x)\,dx\\ &=O\left(\max\left\{Tf(T)e^{-c\sqrt{\log T}},1\right\}\right)+\int_2^{T/2} f'(x) e^{-c\sqrt{\log x}}\,dx \\ & \quad +\int_{T/2}^T f'(x)e^{-c\sqrt{\log x}}\,dx\\ &\ll \max\left\{f(T)e^{-c\sqrt{\log T}},1\right\} \end{align*} Here, I've skipped a few relatively minor details because it's dependent on the specific function $f$ that you are summing. First: the constant $c$ in the exponent changes (but remains bounded below by a positive constant). The split of the integral in the second step is not always strictly necessary, but there are instances where it is beneficial, so I wanted to write it down here. Lastly, the first integral in the second step has not been evaluated whereas the second integral can be bounded above using the lower bound $T/2$ for the exponential term and the fundamental theorem of calculus.
Applying the Approach to this Problem
In this problem, we have $f(p)=\frac{\log p}{p}$, or in the continuous analog, $f(x)=\frac{\log x}{x}$. Inserting this into the integral, we need to calculate the main term $$ \int_2^T\frac{\log x-1}{x\log x}\,dx=\int_2^T\frac{1}{x}-\frac{1}{x\log x}\,dx $$ Integrating, we have the main term equal to $$ \log T-\log\log T+O(1) $$ I'll leave verifying the error terms for you, but this shows the following asymptotic: $$ \sum_{p\leq x\\ p\text{ prime}}\frac{\log p}{p}\sim \log x $$
Riemann's Hypothesis
One of the reasons RH is frequently assumed in analytic number theory papers is because the estimate for primes becomes the much better $$ \pi(x)=\int_2^x {\frac{{dt}}{{\log t}}}+O(\sqrt{x}\log x). $$