Let $F:\mathbb{C}^n\to \mathbb{C}^n$ be a homeomorphism homogeneous of degree 1 (i.e., $F(tx)=tF(x)$, $t>0$) and $g:\mathbb{C}^n\to \mathbb{C}$ a homogeneous polynomial of degree $k$. Let $L$ ($0\in L$) a complex line and $H=F(L)$ such that $g^{1}(0)\cap H=\{0\}$. It is true that $h=g_H\circ F_L$ has topological degree less than or equal to k? This is true if F is linear map!

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 $$ [0,2\pi]\ni \theta \mapsto g\bigl(\, F(e^{i\theta},0,\dotsc, 0)\bigr)\in\mathbb{C}\setminus 0. $$ $\newcommand{\bC}{\mathbb{C}}$ Edit. The computation of the above winding number could be tricky. Here is an example that shows that the topological degree need not be equal to the algebraic degree. Assume $n=2$, $g(z_1,z_2)=z_1$, and $L=\{(z_1,0)\}\subset \bC^2$. Set $$\Sigma=\bigl\{\, (z_1,z_2);\;\;z_1^2+z_2^2 =1\,\bigr\}\subset \bC^2. $$ We will construct degree $1$ homogeneous homeo $\bC^2\to\bC^2$ by fixing a homeo $$\phi: \Sigma\to \Sigma. $$ We get a homeo $F_\phi: \bC^2\to\bC^2$ by setting $$F_\phi(tp)=t\phi(p),\;\;\forall p\in\Sigma,\;\;t>0. $$ Take a tiny unknotted circle $K\subset \Sigma$ contained in a small neighborhood of the point $(1,0)\in \Sigma$. In particular, the projection of $K$ on the first coordinate axis $L$ is a small loop $K'$ with winding number $0$ because $K'$ is contained in a tiny disk of $L$ disjoint from the origin. The circle $\newcommand{\ve}{{\varepsilon}}$ $$ C =L\cap\Sigma=\bigl\{ (z_1,0),\;\;\;z_1=1\,\bigr\}\subset \Sigma. $$ describes another unknotted circle on $\Sigma$. In particular, it is isotopic to the circle $K$ and thus there exists a homeomorphism $\phi: \Sigma\to\Sigma$ such that $\phi(C)=K$. The loop $$ g\circ \phi: C\to \bC\setminus 0 $$ coincides with the loop $K'$ whose winding number is zero. This shows that $$ \deg g\circ F_\phi_L=0\neq \deg g=1. $$ 

