How ROCA get the polynomial used with coppersmith
You are correct—you do want to find a small root of [*]
$$
  f(x) = M \cdot x + (65537^a \bmod M) \bmod N
$$
modulo a divisor $p$ of $N$. In other words, you want to find a divisor of $N$ in the residue class $M \cdot x + (65537^a \bmod M)$. This is explicitly accomplished by the Coppersmith/Howgrave-Graham method, as long as the appropriate restrictions are respected.

However, Coppersmith's method to find such roots traditionally expects a monic polynomial as input. So, to make it monic we simply divide it by the coefficient associated with $x$, that is, $M$:
$$
  g(x) = \frac{1}{M}f(x) = x + \frac{65537^a \bmod M}{M} \bmod N\,.
$$
The roots of $g(x)$ are the same as $f(x)$; to see this decompose the polynomial into its factorization:
$$
   f(x) = (x - \alpha_0)(x - \alpha_1)\dots(x - \alpha_n)\,,
$$
$$
g(x) = \frac{1}{M}(x - \alpha_0)(x - \alpha_1)\dots(x - \alpha_n)\,,
$$
and notice that both have the same roots $\alpha_i$. The Coppersmith method finds the appropriate root modulo a factor of $N$ once you guess the correct $a$.

[*]: The ROCA paper uses alternative $M'$ there, but this is irrelevant here.