Variance of Riemann integral of Stochastic integral

Question

Let $f: \mathbb{R} \to \mathbb{R}$ be deterministic and let $W$ be a standard Brownian motion. Then by Ito's isometry we know $$ Var\left( \int_0^u f(s) dW(s) \right) = \int_0^u f^2(s) ds. $$ Now, consider computing the variance of $$ I := \int_t^T g(u)\left(\int_0^u f(s) dW(s) \right)du = \int_t^T \int_0^u g(u) f(s) dW(s) du. $$ I'm not sure of the technique to use here, but one method I've tried is changing the order of integration to get it in terms of stochastic integrals: $$ I = \int_0^t \int_t^T g(u)f(s)du dW(s) + \int_t^T \int_s^T g(u)f(s)du dW(s) =: I_1 + I_2. $$ Then $$ Var(I) = Var \left(I_1 \right) + Var \left(I_2 \right) + 2Cov \left(I_1, I_2 \right), $$ but I'm unsure about how to compute the covariance term. I suspect it may be zero, since the intervals of integration for the Brownian motions don't overlap, in which case computing the variances would come easy using Ito's isometry. However I'm not sure how to convince myself of this (or even if it's true!) Any thoughts, or better ideas on proceeding with this computation?