Consider the following fiber bundle between smooth manifolds $E'\rightarrow B'$ with a fiber $F'$.
Now let $E$ and $ B$ be compact submanifolds of $E'$ and $B'$ respectively, such that $E\rightarrow B $ is fiber bundle with a compact fiber $F$.
Is there a fiber bundle $F'\rightarrow F$? Is the fiber compact?
First note that in a fiber bundle with nice enough topological spaces for which the total space is compact, the fiber is automatically compact. One version of "nice enough" is "points are closed in the base". This follows because the fiber is the inverse image of a closed subset of the base, so closed, and a closed subset of a compact space is always compact.
With that out of the way, the answer to your question about bundles $F'\rightarrow F$ is, in general, no. Here is an example: begin with the bundle $S^2\rightarrow S^2\times S^2\rightarrow S^2$ where the projection map is projection onto the first factor. Consider the product of equators inside of $S^2\times S^2$, with same projection. This is the bundle $S^1\rightarrow S^1\times S^1\times S^1$.
Thus, $F' = S^2$ and $F = S^1$.
Now, I claim there is no bundle $\pi:F'\rightarrow F$. First, note that the fiber $F''$ must be compact, so it has an Euler characteristic. Then $2 = \chi(F') = \chi(F)\chi(F'') = 0$, a contradiction.