Writing the characteristic function of the primes as $\dfrac{\theta(n) - \theta(n-1)}{\log n}$ is a good start. But one doesn't use the monotonicity of $\theta(n)$, it's the monotonicity of $F(n)/\log n$ that one uses, together with upper and lower bounds for $\theta(n)$. A summation by parts yields
$$\sum_{n = 2}^N \bigl(\theta(n) - \theta(n-1)\bigr)\frac{F(n)}{\log n} = \theta(N) \frac{F(N+1)}{\log (N+1)} + \sum_{n = 2}^N \theta(n)\biggl(\frac{F(n)}{\log n} - \frac{F(n+1)}{\log (n+1)}\biggr) \tag{1}$$
using $\theta(1) = 0$. All terms on the right hand side are non-negative (that's also the case on the left, but we don't use that), and so we make the right hand side smaller/larger if we replace $\theta(n)$ with something smaller/larger. We use Chebyshev's result that there are $c_1, c_2 > 0$ such that
$$c_1\cdot n \leqslant \theta(n) \leqslant c_2\cdot n$$
for all $n \geqslant 2$ and replace $\theta(n)$ with $c_k\cdot n$ on the right hand side of $(1)$. Then reversing the summation by parts gives
\begin{align}
c_kN\frac{F(N+1)}{\log (N+1)} + \sum_{n = 2}^N c_kn\biggl(\frac{F(n)}{\log n} - \frac{F(n+1)}{\log (n+1)}\biggr) &= \sum_{n = 2}^N \bigl(c_k n - c_k(n-1)\bigr) \frac{F(n)}{\log n} \\
&= c_k\sum_{n = 2}^N \frac{F(n)}{\log n}.
\end{align}
Thus we find
$$c_1 \sum_{n = 2}^N \frac{F(n)}{\log n} \leqslant \sum_{p \leqslant N} F(p) \leqslant c_2 \sum_{n = 2}^N \frac{F(n)}{\log n}. \tag{2}$$
From $(2)$ it is clear (since $c_k > 0$) that convergence of $\sum_p F(p)$ implies the convergence of $\sum_{n = 2}^{\infty} \frac{F(n)}{\log n}$ and vice versa.
