Here's the inclusion case when both groups are split (following up on my comment):
Firstly, let's observe a few things. Let $\mathfrak{g} \leq \tilde{\mathfrak{g}}$, $\mathfrak{p} \leq \tilde{\mathfrak{p}}$, $\mathfrak{a} \leq \tilde{\mathfrak{a}}$. Since we are focusing on the split case, we can simply assume $\mathfrak{g}_0 = \mathfrak{a}$, $\tilde{\mathfrak{g}}_0 = \tilde{\mathfrak{a}}$ are Cartan subalgebras. We can quickly see that in this case $\mathfrak{g}_0 = \mathfrak{g} \cap \tilde{\mathfrak{g}}_0$. I'm not sure that holds in more generality though. For a start, consider the degenerate case that $\mathfrak{g}$ is compact when $\mathfrak{p},\mathfrak{a} = 0$ and $\mathfrak{g}_0 = \mathfrak{g}$.
Then we can show:
$$ \bigoplus_{\alpha \in \Sigma} \mathfrak{g}^\alpha = \mathfrak{g} \cap \bigoplus_{\beta\in \tilde{\Sigma}} \tilde{\mathfrak{g}}^\beta. $$
To do this, write $X \in \mathfrak{g}^\alpha$ as $X= \tilde{H} + \sum_{\beta \in \tilde{\Sigma}} X_\beta$ for $\tilde{H} \in \tilde{\mathfrak{g}}_0$. Then take any $H \in \mathfrak{g}_0$ so $[H,X] = \sum_{\beta \in \tilde{\Sigma}} \beta(H)X_\beta$ and $X - \frac{1}{\beta'(H)}[H,X] = \tilde{H} + \sum_{\beta \in \tilde{\Sigma}} \left(1 - \frac{\beta(H)}{\beta'(H)}\right)X_\beta$ for some $\beta' \in \tilde{\Sigma}$ with $\beta(H') \neq 0$. This new thing is still in $\mathfrak{g}$ but we have reduced the terms in the sum and we can keep going recursively until we reach $\tilde{H}$ so that $\tilde{H} \in \mathfrak{g}_0$ but $X \in \mathfrak{g}^\alpha$ then implies $\tilde{H} = 0$. Thus $\bigoplus_{\alpha \in \Sigma} \mathfrak{g}^\alpha \subset \mathfrak{g} \cap \bigoplus_{\beta\in \tilde{\Sigma}} \tilde{\mathfrak{g}}^\beta$ and we have equality by considering dimensions.
In light of that,
$$\sum_{\beta \in \tilde{\Sigma}} \alpha(H) X_\beta = \alpha(H)X = [H,X] = \sum_{\beta \in \tilde{\Sigma}} \beta(H) X_\beta. $$
Therefore for each $X_\beta \neq 0$, $\beta(H) = \alpha(H)$. In particular, any such $\beta$ differ by an element of $\operatorname{ann}(\mathfrak{g}_0) = \{f \in \tilde{\mathfrak{g}}_0^*|f|_{\mathfrak{g}_0} = 0\}$ and at least 1 such $\beta$ must exist.
Now to what this means for the root systems. We have a natural identification of $\mathfrak{g}_0^* \cong \tilde{\mathfrak{g}}_0^*/\operatorname{ann}(\mathfrak{g}_0)$ and thus a quotient map $\pi:\tilde{\mathfrak{g}}_0^* \to \mathfrak{g}_0^*$. Then what we have shown is that $\Sigma \subset \pi(\tilde{\Sigma})$ i.e. that $\Sigma$ is a subset of a quotient of $\tilde{\Sigma}$.
Note that this includes all sorts of possibilities. If $\mathfrak{g}_0 = \tilde{\mathfrak{g}}_0$ we have the maximal rank subalgebras which are given by root subsystems (see here) and more general subsystems. We can also find foldings of Dynkin diagrams and even the $B_3 \to G_2$ pseudo-folding (as discussed here).
The only thing which I don't quite know is how to classify what kinds of quotients are acceptable. Note that the quotient doesn't have to be reduced here (even though both our groups are split) as in the case Torsten mentions of $\operatorname{SO}(p,p+1) \subset \operatorname{SL}(2p+1)$ where the quotient is of type $BC_p$ and we take the $B_p$ inside it.
Note, moreover, that we can conclude that for the root hyperplanes we have $M_\alpha = \tilde{M}_\beta \cap \mathfrak{g}_0$, for each $\beta$ with $\pi(\beta) = \alpha$.
If there is a Weyl group relationship I would expect it to be the other way round. Something like $\psi:\tilde{W} \to W$ with $\psi(s_\beta) := s_{\pi(\beta)}$ (with $\psi(s_\beta)= \mathrm{id}$ if $\pi(\beta) = 0$, etc.). But I cannot see if this defines a valid homomorphism.
Edit: I'm not sure how well this argument would extend to the non-split case as it does depend on the property $\mathfrak{g}_0 = \mathfrak{g} \cap \tilde{\mathfrak{g}}_0$ which I am not convinced of in general.
As an example suggested by Torsten, I calculated what some of the quotient relations for $E_6$ from $A_{26}$ would be. Note, I obtained these by first, finding an explicit representation of a Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{e}_6$ as a subalgebra of the usual diagonal Cartan subalgebra $\tilde{\mathfrak{h}}$of $\mathfrak{sl}_{27}$. Then I compared roots of the larger algebra restricted to $\mathfrak{h}$ (actually I computed their values on several randomly chosen elements) and noted which ones were the same. Note that roots in $A_{26}$ are always of the form $\alpha_i + \alpha_{i+1} + \cdots + \alpha_{i+j}$ where $\alpha_i$ are simple roots.
Then denoting by $\alpha \equiv \beta$ the roots agreeing on $\mathfrak{h}$, I obtained the following relations
- $ \alpha_{1} \equiv \alpha_{4} \equiv \alpha_{7}$
- $ \alpha_{2} \equiv \alpha_{5} \equiv \alpha_{8}$
- $ \alpha_{10} \equiv \alpha_{13} \equiv \alpha_{16}$
- $ \alpha_{11} \equiv \alpha_{14} \equiv \alpha_{17}$
- $ \alpha_{19} \equiv \alpha_{22} \equiv \alpha_{25}$
- $ \alpha_{20} \equiv \alpha_{23} \equiv \alpha_{26}$
- $ \alpha_{8} + \alpha_{9}+ \alpha_{10}+ \alpha_{11} \equiv \alpha_{17} + \alpha_{18}+ \alpha_{19}+ \alpha_{20}$
- $ \alpha_{6} + \cdots + \alpha_{10} \equiv \alpha_{15} + \cdots + \alpha_{19}$
- $ \alpha_{3} + \cdots + \alpha_{9} \equiv \alpha_{15} + \cdots + \alpha_{21}$
- $ \alpha_{2} + \cdots + \alpha_{9} \equiv \alpha_{17} + \cdots + \alpha_{24}$
- $ \alpha_{7} + \cdots + \alpha_{14} \equiv \alpha_{16} + \cdots + \alpha_{23}$
- $ \alpha_{4} + \cdots + \alpha_{12} \equiv \alpha_{17} + \cdots + \alpha_{25}$
- $ \alpha_{1} + \cdots + \alpha_{15} \equiv \alpha_{3} + \cdots + \alpha_{9}$
- $ \alpha_{1} + \cdots + \alpha_{13} \equiv \alpha_{2} + \cdots + \alpha_{10}$
This seems like enough relations but I don't know if there is some redundancy. I certainly didn't do an exhaustive search. Note there is a very distinct pattern in the first ones but the later ones are less clear to me. Also I've written these all as positive roots and mostly as relations between sums of the same length (because of the way I found them) but possibly there are better ways to write these.
