You cannot prove that $\diamondsuit(E\cap F)$ follows from $\diamondsuit(E)$ and $\diamondsuit(F)$, even assuming that $E\cap F$ is stationary. (Of course, if $E\cap F$ is not stationary, then $\diamondsuit(E\cap F)$ fails.)
The reason is that it is consistent to have disjoint stationary sets $A,B,C$ with $\diamondsuit(A)$, $\diamondsuit(B)$ and $\lnot\diamondsuit(C)$. But then you can take $E=A\cup C$, $F=B\cup C$.
That the set you call $I$ is a normal ideal is a result of Shelah (in his famous "Whitehead groups may be not free, even assuming CH. I", Israel J. Math. 28 (1977), no. 3, 193–204). The result holds for any $\kappa$ such that $\diamondsuit_\kappa$ holds, not just for $\kappa=\omega_1$.
The argument (for $\omega_1$) is as follows: (I use sets rather than characteristic functions, but the translation should be straightforward.) First, since $\diamondsuit$ holds, $\omega_1\notin I$. Obviously, if $A\in I$ and $B\subseteq A$, then $A\in I$. To prove that $I$ is normal (which also gives us that it is closed under finite unions, though this of course can be seen directly), suppose $\diamondsuit(E)$ holds, and $f:E\to\omega_1$ is regressive. We need to see that for some $\alpha$, $\diamondsuit(f^{-1}(\{\alpha\}))$ holds.
Start by noting that there is a sequence $(S_\alpha\mid\alpha<\omega_1)$ with $S_\alpha\subseteq\alpha\times\alpha$ and such that for any $X\subseteq\omega_1\times\omega_1$, we have that $\{\alpha\in E\mid X\cap(\alpha\times\alpha)=S_\alpha\}$ is stationary.
(This is easy: Fix a bijection $b:\omega_1\times\omega_1\to\omega_1$ and use it to "pull back" your original diamond sequence. Kunen has several variations of this idea.)
Ok, now, for $\alpha,\beta<\omega_1$, let $$S^\beta_\alpha=\{\zeta<\alpha\mid(\zeta,\beta)\in S_\alpha\}.$$ Check that for some $\beta$, the sequence $(S^\beta_\alpha\mid\alpha<\omega_1)$ is a $\diamondsuit(f^{-1}(\{\beta\}))$ sequence.
For this, assume otherwise, so for each $\beta$ there is a set $X_\beta\subseteq\omega_1$ that is not guessed correctly stationarily often, so there is a club $C_\beta$ such that $X_\beta\cap\alpha\ne S^\beta_\alpha$ whenever $\alpha\in C_\beta\cap f^{-1}(\{\beta\})$. Let $$X=\bigcup_\beta X_\beta\times\{\beta\}$$ and let $C$ be the diagonal intersection of the $C_\beta$. Then it follows just unraveling definitions, that $X\cap(\alpha\times\alpha)\ne S_\alpha$ whenever $\alpha\in C\cap E$. This contradicts that the sequence of $S_\alpha$ was a diamond sequence on $E$, and we are done.
As for the second question, we can in fact show that the following implies diamond:
There is a sequence $(S_\alpha\mid\alpha<\omega_1)$ such that $S_\alpha$ is a countable subset of ${\mathcal P}(\alpha)$ for each $\alpha$, and whenever $X\subseteq\omega_1$, then $X\cap\alpha\in S_\alpha$ for some infinite $\alpha$.
There are several approaches to this fact. It is not complicated, and you may benefit from working through the details. I recommend you take a look at Devlin's "Constructibility" book, where this is stated as an Exercise (in III.3) and a generous hint is provided. Or you can look at the paper where it originally appeared, "Variations on $\diamondsuit$", by Devlin, The Journal of Symbolic Logic, Vol. 44, No. 1, (Mar., 1979), pp. 51-58.
Kanamori wrote years ago a small survey on diamond that was to be part of the second volume of his book on large cardinals. I imagine if you email him he will send you a copy, it was to be part of Chapter VII, "Higher combinatorics". For a truly excellent and up to date survey on diamond together with references, see "Jensen's diamond principle and its relatives", by Assaf Rinot. The paper appeared in "Set Theory and Its Applications", a volume published by the AMS that I co-edited, and can be obtained from Assaf's page.
The idea is very close to forcing. How would you add a generic diamond sequence? You will approximate it by initial segments.
How would you do it over $L$? You approximate it using the $<_L$ order.
The key point is that given a countable model, $M$, of $\sf ZF-P$, we can diagonalise over the sets that it knows to produce a counterexample. If you prefer, think about this as just avoiding all the countably many countable sets that are in the model. Easy, I know.
Now, since every subset of $\omega_1$, including the diamond sequence itself (which formally is a subset of $\omega_1\times\omega_1$), appears in $L_{\omega_2}$, if the definition didn't work, this would be already true in $L_{\omega_2}$, and we have a definable $<_L$-minimal counterexample. Say $(A,C)$ where $A$ is a set and $C$ is a club.
Take $M\prec L_{\omega_2}$, then by definability our sequence, as well our counterexample, are all in $M$. Letting $L_\gamma$ be the transitive collapse of $M$ and letting $\delta$ be the ordinal $\omega_1$ is mapped to, namely, $\delta=\omega_1^{L_\gamma}$, we get that $(A\cap\delta,C\cap\delta)$ is exactly the image of $(A,C)$ under the isomorphism, and that the image of the sequence $(A_\alpha\mid\alpha<\omega_1)$ is just its truncation to $\delta$.
This means that in $L_\gamma$, this is the $<_L$-minimal counterexample, and therefore in $L_{\omega_2}$ this is also the $<_L$-minimal counterexample for that stage, which meant that we took that $A\cap\delta$ to be $A_\delta$. And that is a contradiction.
As I mention above, it is much easier to understand this claim if you understand forcing. This is quite literally "constructing the generic using $<_L$-minimal counterexamples". And if you really understand forcing, there is a method to force $\sf PFA$ and other forcing axioms by "least rank counterexample" that kind of operates on the same principle (since it replaces adding a Laver diamond sequence as a first step).
Best Answer
To see that $C$ is a club, observe that for limit ordinals $\alpha$: $$ M_{A,\alpha} = \bigcup_{\beta < \alpha} M_{A,\beta} $$ Then:
It is certainly closed, as if $M_{A,\alpha_n} \cap \omega_1 = \alpha_n$, and $\alpha = \sup{\alpha_n}$, then: $$ \alpha = \bigcup_{n \in \omega} \alpha_n = \bigcup_{n \in \omega} M_{A,\alpha_n} \cap \omega_1 = M_{A,\alpha} \cap \omega_1 $$
To see that it is unbounded, let $\alpha < \omega_1$, and let $\sigma_0 > \alpha$ be such that $M_{A,\sigma_0} \cap \omega_1 > \sigma_0$ (if no such $\sigma_0$ exists, then unboundedness follows immediately). Now define recursively $\sigma_{n+1} := M_{A,\sigma_n} \cap \omega_1$. If the sequence becomes constant eventually, then we are done. Otherwise, consider $\sigma = \sup_n \sigma_n$. Then $\sigma$ is a limit ordinal, and we have that $M_{A,\sigma} \cap \omega_1 = \sigma$.