To avoid complicating the maths with vector notation, instead consider just a 2-day return as opposed to a 30-day.
Then your two, 2-day, series can be stated as ($X_t, Y_t$) where $X_t=x_t+x_{t-1}$ and $Y_t=y_t+y_{t-1}$, for little $x,y$ the daily, i.e. 1-day (log) returns.
Then,
$$Cov(X_t,Y_t) = E[(X_t-\mu_X)(Y_t-\mu_Y)] = E[(x_t-\mu_x+x_{t-1}-\mu_x)(y_t-\mu_y+y_{t-1}-\mu_y)]$$,
which after expanding out gives you four terms,
$$Cov(X_t,Y_t)=E[(x_t-\mu_x)(y_t-\mu_y)+(x_{t-1}-\mu_x)(y_{t-1}-\mu_y)]+E[(x_t-\mu_x)(y_{t-1}-\mu_y)+(x_{t-1}-\mu_x)(y_{t}-\mu_y)]$$
$$Cov(X_t,Y_t)=2Cov(x_t,y_t) + Cov(x_{t-1},y_t) + Cov(x_t,y_{t-1})$$
In other words the covariance of the 2-day time frame is twice the covariance of the daily time frame plus the time offset covariances. In option theory and Brownian motion these time offset terms are set to zero so that the covariance of an n-long period is simply n times the covariance of a single period.
If you aspire to the same modelling characteristics in your data then you can use daily returns and multiply them by 30 as the estimator for your 30-day covariance.