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 $Z_g:=g^{-1}(0)$ at a point $p_0\neq 0$, then the  the intersection $Z_g\cap F(L)$ contains the whole ray $tp_0$, $t\geq 0$ and thus $g\circ F|_L$ 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 $Z_g\cap F(L)=\lbrace 0\rbrace$.  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

$$ [0,2\pi]\ni \theta \mapsto g\bigl(\, F(e^{i\theta},0,\dotsc, 0)\bigr)\in\mathbb{C}\setminus 0. $$

$\newcommand{\bC}{\mathbb{C}}$