We may assume that $\phi'$ is standard. Let $a\in A$ be nonzero. Because ring homomorphisms map units to units, $\phi_a'$ and $\phi_a$ are standard of the same degree. Hence $\phi'$ and $\phi$ have the same rank.

At this point of the proof, Laumon assumes that $\phi$ is standard—I'll do so as well. Let $a\in A$ be nonzero. Then

$$ \phi_a = \sum_i \phi_{a,i} \tau^i$$

with $\phi_{a,-d\deg(\infty)\infty(a)} \in k^{\times}$ and $\phi_{a,i} = 0$ in higher degrees. Note that an element $x$ of $\mathcal{O}$ is a unit if and only if is a unit mod $\mathfrak{m}$, because $\mathfrak{m}$ consists of nilpotents. Hence

$$ \phi'_a = \sum_i \phi'_{a,i} \tau^i$$

with $\phi'_{a,-d\deg(\infty)\infty(a)} \in \mathcal{O}^{\times}$ and $\phi'_{a,i} \in \mathfrak{m}$ in higher degrees. Compare this with Definition 1.2.1 to see that $\phi'$ is a Drinfeld module of rank $d$ over $\mathcal{O}$.

We may assume that $\phi'$ is standard. Let $a\in A$ be a nonzero nonunit. Since $\mathfrak{m}$ is a proper ideal, the leading coefficient of $\phi'_a$ cannot vanish mod $\mathfrak{m}$. Thus $\phi_a'$ and $\phi_a$ have the same degree, so $\phi'$ and $\phi$ have the same rank.

We may assume that $\phi'$ is standard. Let $a\in A$ be nonzero. Because ring homomorphisms map units to units, $\phi_a'$ and $\phi_a$ are standard of the same degree. Hence $\phi'$ and $\phi$ have the same rank.