To understand correlation in HW2F (or G2++) model it suffices to compute the correlation for the log of bonds at two different maturities. Your intuition is right, that its correlation is not just driven by the correlation of the two Brownian motion, but also by their mean reversions. The model is still different from HW1F as long as the two mean reversions are different.
Let us recall the G2++ model dynamics:
$$dr_t=\theta_t^\prime+dx_t+dy_t$$
$$dx_t=-ax_t dt +\sigma dW_t^x$$
$$dy_t=-bx_t dt +\eta dW_t^y$$
$$d\langle W^x,W^y\rangle_t=\rho dt$$
When $\rho=-1$, we have $dr_t=[\theta_t^\prime-(ax_t+by_t)]dt+(\sigma+\eta)dW_t^x$, which will be different than $x_t+y_t$ provided $a\ne b$.
The term with the smallest mean reversion will indeed be more volatile than the other one. This is theoretically justified by the expression of the correlation:
$$Corr(x_t,y_t )=\frac{\sqrt{ab}(1-e^{-(a+b)t})}{(a+b)\sqrt{(1-e^{-2at})(1-e^{-2bt})}}.$$
Now, we can focus on calculating the correlation between the long-term and short-term rates. Let us consider the short and long term rates expressed in terms of log ZCBs: $P(t,T)=\exp{r(t,T)(T-t)}$, and let's compute the correlation between $P(t,T_1)$ and $P(t,T_2)$, with $T_1<T_2$.
$$\ln(P(t,T))=-\int_t^T θ_sds-\frac{1-e^{-a(T-t)}}{a}x_t-\frac{1-e^{-b(T-t)}}{b}y_t+\frac{1}{2} V(t,T)$$
with $V(t,T)$ being the conditional variance of the short-rate,
\begin{align}
V(t,T) &= \frac{\sigma^2}{a^2}\left[(T-t)-2\frac{1-e^{-a(T-t)}}{a}+\frac{1-e^{-2a(T-t)}}{2a}\right]\\
&+\frac{\eta^2}{b^2}\left[(T-t)-2\frac{1-e^{-b(T-t)}}{b}+\frac{1-e^{-2b(T-t)}}{2b}\right]\\
&+2\rho\frac{\sigma\eta}{ab}\left[(T-t)-\frac{1-e^{-a(T-t)}}{a}-\frac{1-e^{-b(T-t)}}{b}+\frac{1-e^{-(a+b)(T-t)}}{a+b}\right]
\end{align}
Conditionally on the time $\tau$ filtration, $x_t$ and $y_t$ are both normally distributed stochastic processes, whilst the other two terms are deterministic. So, we can study the covariance of the two bonds with different maturity and obtain:
\begin{align}
Cov_\tau(\ln P(t,T_1),\ln P(t,T_2))&=\mathbb{E}_\tau\left[\left(\frac{1-e^{-a(T_1-t)}}{a}\sigma\int_\tau^t e^{-a(t-u)}dW_u^x+\frac{1-e^{-b(T_1-t)}}{b}\eta\int_\tau^te^{-b(t-u)}dW_u^y\right)\left(\frac{1-e^{-a(T_2-t)}}{a}\sigma\int_\tau^te^{-a(t-u)}dW_u^x+\frac{1-e^{-b(T_2-t)}}{b}\eta\int_\tau^te^{-b(t-u)}dW_u^y\right)\right]\\
&=\frac{(1-e^{-a(T_1-t)})(1-e^{-a(T_2-t)})}{2a^3}\sigma^2(1-e^{-2a(t-s)})\\
&+\frac{(1-e^{-b(T_1-t)})(1-e^{-b(T_2-t)})}{2b^3}\eta^2(1-e^{-2b(t-s)}) \\
&+\frac{(1-e^{-a(T_1-t)})(1-e^{-a(T_2-t)})+(1-e^{-b(T_1-t)})(1-e^{-b(T_2-t)})}{ab(a+b)}\rho\sigma\eta(1-e^{-(a+b)(t-s)}).
\end{align}
$$\Rightarrow \left(\varrho_{(\ln P(t,T_1),\ln P(t,T_2))}\right)_\tau=\frac{Cov_\tau(\ln P(t,T_1),\ln P(t,T_2))}{\sqrt{V(t,T_1)V(t,T_2)}}$$
At this point you can make some simplifications and realise that regardless of the value of $\rho$, the numerator and denominator of $\left(\varrho_{(\ln P(t,T_1),\ln P(t,T_2))}\right)_\tau$ cancel out if $a=b$.
You can now use the Libor-Bond relationship to get the correlation between 3M and 6M libors.
