Prime reciprocal sum \begin{align}
&f(k)=
\begin{cases}
1/k\text{ if $k$ is prime}\\
0\text{ otherwise}\\
\end{cases}
\\
\\
&\sum_{k=1}^{n}f(k)\sim\log\log n+M
\end{align}
where $M$ is the Meissel–Mertens constant, seems far closer to the truth than $$\sum_{{k=1}}^{n}1/p_k\sim\log\log n+M$$
Is this the case?

The plot is of $\sum_{k=1}^{n}f(k)$ (blue )$\log\log n+M$ (red). This is different to this plot
Compare:
singleNumPa[om_, number_] := 
If[number == 1, 0, If[PrimeOmega[number] < om, 1, 0]]/number
Show[Accumulate[singleNumPa[2, #] & /@ Range@1000] // ListLinePlot, 
Plot[Log@Log[x] + .2614972128476427837554268386086958590516, {x, 2, 
1000}, PlotStyle -> Red], PlotRange -> All]
Show[Accumulate[1/Prime[#] & /@ Range@1000] // ListLinePlot, 
ListLinePlot[
Table[(Log@Log[x] + .2614972128476427837554268386086958590516), {x, 
1, 1000}], PlotStyle -> Red], PlotRange -> All]

Update
Apologies to all - misread Wiki article as prime reciprocals as opposed to $p<n$
 A: Using the estimate $\pi(n)=\frac{n}{\log(n)}+O\left(\frac{n}{\log(n)^2}\right)$, your sum can also be written as
$$
\begin{align}
\sum_{k=2}^nf(k)
&=\sum_{k=2}^n\frac{\pi(k)-\pi(k-1)}{k}\\
&=\sum_{k=2}^n\frac{\pi(k)}k-\sum_{k=1}^{n-1}\frac{\pi(k)}{k+1}\\
&=\sum_{k=2}^n\frac{\pi(k)}{k(k+1)}+\frac{\pi(n)}{n+1}\\
&=\sum_{k=2}^n\frac{\frac{k}{\log(k)}+O\left(\frac{k}{\log(k)^2}\right)}{k(k+1)}+\frac{\pi(n)}{n+1}\\[6pt]
&=\sum_{k=2}^n\frac1{(k+1)\log(k)}+O(1)\\[12pt]
&=\log(\log(n))+O(1)
\end{align}
$$
However
$$
\begin{align}
\sum_{k=2}^nf(k)
&=\sum_{k=1}^{\pi(n)}\frac1{p_k}\\
\end{align}
$$
So the difference
$$
\begin{align}
\sum_{k=1}^n\frac1{p_k}-\sum_{k=2}^nf(k)
&=\sum_{k=1}^n\frac1{p_k}-\sum_{k=1}^{\pi(n)}\frac1{p_k}\\
&=\sum_{k=\pi(n)+1}^n\frac1{p_k}\\
&=\sum_{k=\pi(n)+1}^n\frac1{k\log(k)+O(k)}\\[6pt]
&=\sum_{k=\pi(n)+1}^n\frac1{k\log(k)}+O\left(\frac1{\log(n)}\right)\\[6pt]
&=\log(\log(n))-\log(\log(n/\log(n)))+O\left(\frac1{\log(n)}\right)\\[12pt]
&=\log(\log(n))-\log(\log(n)-\log(\log(n)))+O\left(\frac1{\log(n)}\right)\\[9pt]
&=\frac{\log(\log(n))}{\log(n)}+O\left(\frac1{\log(n)}\right)\\[9pt]
&\to0
\end{align}
$$
Therefore, the Meissel-Mertens constant will be the same either way.
A: Call your first sum $S_1(n)$ and the second $S_2(n)$.  Since $p_n\approx n\ln n$, then we have $S_2(n)\approx S_1(n\ln n)$, or alternately $S_1(n)\approx S_2(\frac n{\ln n})$.  But $\ln(\ln(\frac{n}{\ln n}))$ $= \ln(\ln n-\ln\ln n)$ $= \ln(\ln n\cdot(1-\frac{\ln\ln n}{\ln n}))$ $=\ln\ln n+\ln(1-\frac{\ln\ln n}{\ln n})$ $\approx \ln\ln n-\frac{\ln \ln n}{\ln n}$ — and note that I haven't tried to roll any specific constants in here.  The latter value is small and tends to zero as $n\to\infty$, so asymptotically there's no difference between the two expressions.
