A cone is simply a triangle rotated about an axis to make a solid. We'll place it on the $y$-axis instead of the $x$-axis because it looks better standing upright. Now if you slice a smaller cone off the top, then you're left with a conical frustum. This new solid is now the same as an isosceles trapezoid rotated about an axis down the middle. Therefore, if we take a look at an intermediate slice of the intersection of two frustums, we'll get essentially one of four possible situations (up to otherwise superfluous orientation and proportion):

You can see how in the right two cases, the intersection is two conical frustums glued together, base to base, while in the left two cases the intersection is just another frustum. Do keep in mind, however, that the two main frustums don't necessarily look "upside-down" relative to each other - but they will make one of these situations, regardless. (Actually, I'm lying: there's a fifth possible case where one frustum completely contains the other, but the problem is trivial then... and makes my nice picture less symmetrical.) Instead of going over all four possible situations, I'll only lay out the two tools you need in order to do the problem yourself, hopefully. The first is this:
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If you know the top has length (diameter for the frustum) $A$, and you know the bottom has length $B$, and you know the purple line is $u$ distance down from the top and $v$ distance up from the bottom, then the purple line has length equal to the weighted average $(Au+Bv)/(u+v)$. You can use this to find the base diameters of the intersection frustum based on the top and bottom diameters of the main two frustums.
Lastly, a frustum with top diameter $A$, bottom diameter $B$, and height $h$, has volume that can be calculated with the formula $\frac{\pi h}{12}(A^2+AB+B^2)$. If you don't understand how to resolve your intersecting frustums problem now, tell me where you're stuck and I'll try to elaborate.