For (2), the answer is no, not in general. Here is a simple example: take $K_1=\mathbb{Q}(i)$ and $K_2=\mathbb{Q}(\sqrt{-5})$. Then both $K_1/\mathbb{Q}$ and $K_2/\mathbb{Q}$ are (totally) ramified at $p=2$ and $K_1\cap K_2=\mathbb{Q}$, but $F=K_1K_2$ is not totally ramified at $2$. In other words, the extension $\mathbb{Q}(\sqrt{-5},i)/\mathbb{Q}(\sqrt{-5})$ is unramified at $2$ (in fact, $F$ is the Hilbert class field of $K_2$, so it is unramified everywhere).

For (2), the answer is no, not in general. Here is a simple example: take $K_1=\mathbb{Q}(i)$ and $K_2=\mathbb{Q}(\sqrt{-5})$. Then both $K_1/\mathbb{Q}$ and $K_2/\mathbb{Q}$ are (totally) ramified at $p=2$ and $K_1\cap K_2=\mathbb{Q}$, but $F=K_1K_2$ is not totally ramified at $2$. In other words, the extension $\mathbb{Q}(\sqrt{-5},i)/\mathbb{Q}(\sqrt{-5})$ is unramified at $2$ (in fact, $F$ is the Hilbert class field of $K_2$, so it is unramified everywhere).

On the other hand, if you take $K_1=\mathbb{Q}(i)$ and $K_2=\mathbb{Q}(\sqrt{2})$, then $F=K_1K_2$ is totally ramified at $2$ over $\mathbb{Q}$. (Here $F=\mathbb{Q}(\zeta_8)$.)

In both cases, $I_1=D_1=G_1$ and $I_2=D_2=G_2$ (in your notation) but in the first case the inertia in the compositum has order $2$ and in the second case it has order $4$. This shows that one needs to know more than the decomposition and inertia subgroups at a prime in each $K_i$ to understand the ramification index in the compositum.