The multiplication of rank for finite projective modules 
Assuming all rings are commutative, suppose $M$ is finite projective module over $R$ of rank $m$ which is itself an $R'$-algebra that is as a module finite projective of rank $n$. How can we show that $M$ is of rank $mn$ over $R'$?

Where the rank means localized at any prime, the module is free of that rank.
It is straightforward that $M$ is finite projective but I have difficulty because even though primes of $R'$ lie below primes of $R$, $M$ localized at the two primes is different.
This is question 2 in section 7.9 of Jacobson's Basic Algebra II.
 A: Let me use a slightly different notation: We have $R' \to R \to M$, where $R$ is $R'$-projective of rank $s$ and $M$ is $R$-projective of rank $r$.
The previous exercise states that 

if $M$ is $R$-projective of rank $r$, then for any maximal ideal $m$ of $R$, $\dim_{R/m} M/mM = r$.

Since you already showed that $M$ is $R'$-projective, it suffices to show that for a maximal ideal $m$ of $R'$, $\dim _{R'/m} M/mM = sr$.
The extension $R'/m \to R/ mR$ is integral and $R'/m$ is a field.
Hence $R / mR$ is a finite product of Artinian local rings, and $M /mM$ is a module over this product. Write 
$$R/ mR \cong R_1 \times R_2 \times \cdots \times R_q.$$ 
Here, each $R_i$ is a localization of $R / mR$ at a maximal ideal, and $R_i$ is finite over $R'$.
We conclude that 

$\dim_{R'/mR'} R/mR = \sum \dim_{R'/mR'} R_i = s$.

Now, we apply the exercise to $R$ and $M$. As $M$ is $R$-projective, for each maximal ideal $n$ of $R$, $\dim_{R/n} M/ nM = r$.
Let $n_i$ be the maximal ideal which is the kernel of the morphism
$$
R \to R/ mR \cong R_1 \times R_2 \times \cdots \times R_q \to R_i.$$ 
Then $(R/mR)/ n_i(R/mR) \cong R/n_i R$, and $\dim_{R/{n_i}} M/{n_i}M = r$.
Lastly, we these numbers together. We have that 
$M/mM \cong \oplus (R/{n_i})^r$, and 
$$\dim_{R'/m} M/mM 
= \sum r \dim_{R'/m} R_i = r \sum \dim_{R'/m} R_i = r \dim_{R'/m} R/mR = rs.$$
