Your question needs to be phrased more carefully.
Take $n=2$, $g(z_1,z_2)=z_1z_2$, $F=$ the identity map $\mathbb{C^2}\to\mathbb{C^2}$ and $L$ is the subspace
$$L=\lbrace(z_1,0),\;\;z_1\in\mathbb{C}\rbrace\subset \mathbb{C}^2. $$
Then $g\circ F|_L=0 $ and $g\circ F|_L$ does not have a degree.
In general for a map $g: \mathbb{C}\to \mathbb{C}$ to have a degree it has to be a proper map. The zero set of a homogeneous polynomial is a conical set in $\mathbb{C}^n$ and so is the image $F(L)$. If $F(L)$ intersects $g^{-1}(0)$ at a point $p_0\neq 0$, then the the intersection $g^{-1}()\cap F(L)$ contains the whole ray $tp_0$, $t\geq 0$ and thus $g\circ F$ cannot be proper and thus it does not even have a well defined degree.
It is not hard to see that $g\circ F|_L$ is proper if and only if $g^{-1}(0)\cap F(L)=\lbrace 0\rbrace$. Assume that $L$ is the first coordinate line in $\mathbb{C}^n$. The degree of $g\circ F|_L$ is then the winding number of the loop
$$ [0,2\pi]\ni \theta \mapsto g\bigl(\, F(e^{i\theta},0.\dotsc, 0)\bigr)\in\mathbb{C}\setminus 0. $$