In general, just knowing the holonomy of $M$ will not tell you much about the holonomy of $\tilde M$. For example, if $M$ is isometric to the sphere of radius $r>0$ (and the dimension of $M$ is at least $2$), then the holonomy of $M$ is $\mathrm{SO}(n)$. When $r\not=1$, the holonomy of $\tilde M$ will be $\mathrm{SO}(n{+}1)$, but when $r = 1$, the holonomy of $\tilde M$ will be trivial, because $\tilde M$ will be isometric to $\mathbb{R}^{n+1}$.

As another example, when $M^6$ is strictly nearly K\"ahler, then the holonomy of $M$ will be $\mathrm{SO}(6)$. In most cases, the holonomy of $\tilde M$ will be $\mathrm{SO}(7)$, but, if the scalar curvature is the right constant (I forget the exact value), the holonomy of $\tilde M$ will be $\mathrm{G}_2$ or trivial.

In general, you need to know more about the metric on $M$ than just its holonomy in order to compute the holonomy of $\tilde M$. Usually, the metric on $M^n$ has to be quite special in order for the holonomy of $\tilde M$ to be a proper subgroup of $\mathrm{SO}(n{+}1)$.

There might be some information on this topic in Besse's *Einstein Manifolds*, particularly the chapter on holonomy. Otherwise, you could look at some papers, for example, the recent papers of Mark Haskins and his collaborators on metrics on cones with special holonomy. I'm sorry, but I'm travelling and don't have access to those references right now.

**Answer to the added specific question:** Yes, that's correct, the cone on a $\mathrm{G}_2$-holonomy metric *never* has holonomy $\mathrm{Spin}(7)$. On the other hand, the cone on a strictly nearly $\mathrm{G}_2$-structure (one that has the right scalar curvature and is not the $6$-sphere) will have holonomy $\mathrm{Spin}(7)$