As Lucia says, in order to estimating $\sum_{n\le x} v^{\omega(n)}$ the key phrase is Selberg-Delange method (there is a chapter about it in Tenenbaum's book "Intro to analytic and probabilistic number theory").

As far as estimating $S(x) =\sum_{n\le x} \omega(\omega(n))$ is concerned, which is what you actually want to do, it is be possible to apply a more direct approach, as Lucia also indicates. I shall elaborate on this below. (Note that this approach is more direct only on the surface, as it depends on results derived via the Selberg-Delange method.) The starting point is the identity
$$
S(x) = \sum_{r\ge1} \omega(r) \pi_r(x) ,
$$
where
$$
\pi_r(x)=\#\{n\le x: \omega(n)=r\}.
$$
Now, we know that
\begin{align*}
\pi_r(x)
&=\left( \lambda\left(\frac{r-1}{\log\log x} \right) + O\left(\frac{r}{(\log\log x)^2}\right)\right)\frac{x}{\log x} \frac{(\log\log x)^{r-1}}{(r-1)!} \\
&\asymp \frac{x}{\log x} \frac{(\log\log x)^{r-1}}{(r-1)!},
\end{align*}
uniformly for $r\le 10\log\log x$, where
$$
\lambda(z) = \frac{1}{\Gamma(z+1)} \prod_p\left(1+\frac{z}{p-1}\right)\left(1-\frac1p\right)^{z}
$$
(This is Theorem 4 in page 205 of Tenenbaum's book, 1995 english edition). Larger $r$ should contribute very little to $S(x)$. One can use the Hardy-Ramanujan theorem to bound their contribution, which says that
$$
\pi_r(x) \le \frac{Ax}{\log x} \frac{(\log\log x+B)^{r-1}}{(r-1)!},
$$
for two absolute constants $A$ and $B$ (and all $r\ge1$ and $x\ge3$). So we essentially have that
\begin{align*}
S(x)
&\sim \sum_{r\le 10\log\log x} \omega(r) \pi_r(x) \\
&\sim \sum_{r\le 10\log\log x} \omega(r) \lambda\left(\frac{r-1}{\log\log x} \right) \frac{x}{\log x} \frac{(\log\log x)^{r-1}}{(r-1)!} .
\end{align*}
Most of the contribution to this sum should come from integers $r=\log\log x+O(\sqrt{\log\log x})$ which have about $\log\log\log\log x$ prime factors. Also, for such $r$ we have that $\lambda((r-1)/\log\log x)\sim1$. The sum of $(\log\log x)^{r-1}/(r-1)!$ over such $r$ should be $\sim \log x$ (this is a baby version of the Central Limit Theorem). So it seems that $S(x)\sim x\log\log\log\log x$.

You might be able to obtain this result using your initial approach too, you have to be careful about the dependence of your estimates on $p$ though.