What is the limiting distribution of $\sqrt{n}(\bar{X}_n-p)-\sqrt{n}(\bar{Y}_n-q)$? For two sequence of random variables $X_1,\dots, X_n\sim_{iid} Beroulli(p) X_i$ and $Y_1,\dots, Y_n\sim_{iid} Beroulli(q)$ ($X_i$, $Y_j$ are independent), we have CLT
$$
\sqrt{n}(\bar{X}_n-p)\to N(0,p(1-p))
$$
and
$$
\sqrt{n}(\bar{Y}_n-q)\to N(0,q(1-q))
$$
Using Delta method, we get
$$
\sqrt{n}(\log(\frac{\bar{X}_n}{1-\bar{X}_n})-\log (\frac{p}{1-p}))\to N(0, \frac{1}{p(1-p)})
$$
(1) Question: What is the distribution of $\sqrt{n}(\log(\frac{\bar{X}_n}{1-\bar{X}_n})-\log (\frac{p}{1-p})-\log(\frac{\bar{Y}_n}{1-\bar{Y}_n})-\log \frac{q}{1-q}))$?
Do we have $$\sqrt{n}(\log(\frac{\bar{X}_n}{1-\bar{X}_n})-\log (\frac{p}{1-p})-\log(\frac{\bar{Y}_n}{1-\bar{Y}_n})-\log \frac{q}{1-q}))\to N(0, \frac{1}{p(1-p)}-\frac{1}{q(1-q)})?$$ Otherwise, how to get it?
I think I am right.
(2) Question: what is the consistent estimator of $\frac{1}{p(1-p)}-\frac{1}{q(1-q)}$?
 A: You're almost correct. As a reminder, you have to assume that $p, q \neq 0$.
Assume two independent sequences of random variables, $A_n$ and $B_n$, converge in distribution to $A$ and $B$: $A_n \rightarrow A$ and $B_n \rightarrow B$. Then for any real numbers, $c$ and $d$, we have $c \cdot A_n + d \cdot B_n \rightarrow c \cdot A + d \cdot B$.
In your example, we have
$$A_n := \sqrt{n}\{\log(\frac{\bar{X}_n}{1-\bar{X}_n})-\log (\frac{p}{1-p})\}$$
$$B_n := \sqrt{n}\{\log(\frac{\bar{Y}_n}{1-\bar{Y}_n})-\log (\frac{q}{1-q})\},$$
$$A_n - B_n \rightarrow N\{0, \frac{1}{p(1-p)}\} - N\{0, \frac{1}{q(1-q)}\}$$.
You made a mistake in calculating your variance. For independent random variables $A$ and $B$, $var(A - B) = var(A) + var(B)$. So your two sequences converge to $$N\{0, \frac{1}{p(1-p)} + \frac{1}{q(1-q)}\}.$$
I see two ways to get a consistent estimate of the variance,  $\frac{1}{p(1-p)} + \frac{1}{q(1-q)}$.
1. Multivariate Delta Method
Since we're assuming $\bar{X}_n$ and $\bar{Y}_n$ are $\sqrt{n}$ consistent estimators of $p$ and $q$, we know that
$$\sqrt{n} \begin{bmatrix}
\bar{X}_n - p \\ 
\bar{Y}_n - q\ 
\end{bmatrix} \rightarrow N(0, \Sigma), \text{where}$$
$$\Sigma = \begin{bmatrix}
p(1-p) & 0  \\
0 & q(1-q) \\
\end{bmatrix}. $$
You can definitely estimate $\Sigma$ since it's the variance of two independent Binomial random variables. Plug $\hat{p}$ and $\hat{q}$ into $\Sigma$ to get $\hat{\Sigma}$.
Let $h(p ,q) = log(\frac{p}{1-p}) - log(\frac{p}{1-p})$  and let $\nabla h(p, q)$ be the gradient of $h$.
The Delta Method gives the asymptotic distribution and covariance matrix:
$$\sqrt{n} 
\{h(\bar{X}_n, \bar{Y}_n) - h(p, q)\} \rightarrow N \{0, \nabla h(p, q)^T \cdot \Sigma \cdot \nabla h(p, q) \}$$
A consistent estimate of the variance is given by plugging $\hat{\Sigma}$ into the expression for the variance above.
2. Invariance Property of the Maximum Likelihood Estimate (MLE)
The Invariance Property of the MLE states that $\hat{\theta}$ is a MLE of $\theta$, then $f(\hat{\theta})$ is a MLE of $f(\theta)$. So plug $\hat{p} = \frac{1}{n} \sum_i X_i$  and $\hat{q} = \frac{1}{n} \sum_i Y_i$ into $f(p, q) = \frac{1}{p(1-p)} + \frac{1}{q(1-q)}$
A: I'm answering the title question; the question in the body is unclear to me.

Using the rules for a variance operator we have:
$$\begin{align}
\mathbb{V}(\sqrt{n} (\bar{X}_n-p) - \sqrt{n} (\bar{Y}_n-q))
&= \mathbb{V}(\sqrt{n} (\bar{X}_n-p)) + \mathbb{V}(\sqrt{n} (\bar{Y}_n-q)) \\[14pt]
&= n \cdot \mathbb{V}(\bar{X}_n-p) + n \cdot \mathbb{V}(\bar{Y}_n-q) \\[14pt]
&= n \cdot \mathbb{V}(\bar{X}_n) + n \cdot \mathbb{V}(\bar{Y}_n) \\[8pt]
&= n \cdot \frac{p(1-p)}{n} + n \cdot \frac{q(1-q)}{n} \\[6pt]
&= p(1-p) + q(1-q). \\[6pt]
\end{align}$$
Since linear functions of normal random variables are normal random variables, you have the limiting distribution:
$$\sqrt{n} (\bar{X}_n-p) - \sqrt{n} (\bar{Y}_n-q)
\overset{\text{approx}}{\sim} \text{N}(0, p(1-p) + q(1-q)).$$
