I'm using Milnor's definition of degree mod 2 for a smooth map. Specifically, let $f: M \rightarrow N$ be a smooth map with $M$ compact without boundary, $N$ connected, and $M$ and $N$ having the same dimension. Let $y \in N$ be a regular value for $f$. Then the degree mod 2 of $f$ is $\text{deg}_2(f) = |f^{-1}(\{y\})| \pmod 2$.
Let $f: M \rightarrow N$ and $g:N \rightarrow S$ be smooth maps between manifolds whose domains and codomains satisfy the above conditions, i.e. $M$ and $N$ are compact without boundary and $S$ is connected. I want to show that $$\text{deg}_2(g \circ f) = \text{deg}_2(g) \text{deg}_2(f) \pmod 2.$$
I know that if $y \in S$ is a regular value for $g \circ f$, then for all regular points $p \in (g\circ f)^{-1}(\{y\})$, we have that $$ d_p(g \circ f): T_pM \rightarrow T_yS $$ is surjective. By the chain rule, we get $d_p(g \circ f) = d_{y} g \circ d_pf$. From this composition, we see that $d_yg$ is surjective, so $y$ is a regular value for $g$. However, I currently have no reason to believe that any point in $g^{-1}(\{y\})$ is a regular value for $f$.
This is where I hit a snag, so I'm wondering if this is the correct direction and if so, how to proceed from here.