The ramification index is multiplicative in towers. That is:
If $K\subseteq E\subseteq L$ is a tower of extensions, $P$ is a prime in $K$, $Q$ is a prime in $E$ with $Q\cap K=P$, and Q' is a prime in $L$ with Q'\cap E=Q, then letting $e(Q|P)$ be the ramification index of $Q$ over $P$, e(Q'|Q) the ramification index of Q' over $Q$, and e(Q'|P) the ramification index of Q' over $P$, we have have e(Q'|P) = e(Q'|Q)e(Q|P). To see this, you do the factoring: if we factor $P$ into primes in $E$, then $Q$ occurs with exponent $e(Q|P)$; then we factor in $L$, the exponent of Q' will be its exponent in the factorization of $Q$, e(Q'|Q), times the exponent to which $Q$ occurs in that of $P$, $e(Q|P)$, since $(\mathfrak{P}^a)^b = \mathfrak{P}^{ab}$.
If all extensions are Galois (that is, $L$ and $E$ are both Galois over $K$), then the ramification index depends only on $P$ and not on $Q$ or Q', but it's still multiplicative, but here you get the further information that the ramification index also divides the order of the extension.
All of that information it may give you enough leverage to determine the ramification index in $E_i/K$ in some circumstances (if you know the ramification in $L/K$, then the multiplicativity means that knowing the index in $E_i/K$ is equivalent to knowing it in $L/E_i$). But it may not. For example, if you have a biquadratic extension $L/K$ whose Galois group is the Klein $4$-group, with a prime $P$ that has ramification index $2$, then for the two intermediate fields $E_1$ and $E_2$ you could have that $P$ already ramifies in $E_1$, but the ramification index in $L/E_1$ is $1$; or the other way around.
Added. To answer your final question without reference to discriminants:
No rational prime can have ramification degree $4$ in $K=\mathbb{Q}(\sqrt{-5},\sqrt{-1})$: that would require it to ramify in both $\mathbb{Q}(\sqrt{5})$ and in $\mathbb{Q}(\sqrt{-1})$ by the multiplicativity of the ramification degree, but no prime ramifies in both. So the ramification degree of any ramified rational prime is $2$.
Let $L_1 = \mathbb{Q}(\sqrt{5})$, $L_2=\mathbb{Q}(\sqrt{-5})$, $L_3=\mathbb{Q}(\sqrt{-1})$. These are all the intermediate fields of the extension.
Let $p$ be a prime that ramifies in $K$. Let $Q$ be a prime of $K$ lying over $p$, and let $P_i$ be the prime of $L_i$ lying under $Q$. If $I_p$ is the inertia group of $p$, and $L_i$ is the fixed field of $I_p$, then $e(Q|P_i) = e(Q|p) = 2$, and $e(Q|P_j)=1$ with $j\neq i$. Again, multiplicativity of the ramification tells you that $p$ would then need to ramify in both the "other two" subextensions.
In particular, if $L_i = L_2 = \mathbb{Q}(\sqrt{-5})$ (so that there is a prime of $\mathbb{Q}(\sqrt{-5})$ that ramifies in $K$), then $p$ would have to ramify in both $\mathbb{Q}(\sqrt{5})$ and in $\mathbb{Q}(\sqrt{-1})$. But no rational prime ramifies in both; so we can never have the fixed field of $I_p$ equal to $\mathbb{Q}(\sqrt{-5})$, which means $K/\mathbb{Q}(\sqrt{-5})$ must be unramified everywhere, as claimed.