Let $d\in\mathbb N$, $\alpha\in\mathbb N\uplus\{\infty\}$, $\Omega$ be a bounded $d$-dimensional properly embedded $C^\alpha$-submanifold of $\mathbb R^d$ with boundary, $\nu_{\partial\Omega}$ denote the outward-pointing unit normal field on $\partial\Omega$, $$P:C_c^{\alpha-1}(\mathbb R^d,\mathbb R^d)\to C_c^{\alpha-1}(\partial\Omega)\;,\;\;\;\theta\mapsto\left\langle\left.\theta\right|_{\partial\Omega},\nu_{\partial\Omega}\right\rangle$$ and $g\in C_c^{\alpha-1}(\mathbb R^d,\mathbb R^d)'$ with $\ker P\subseteq\ker g$.
I would like to conclude that there is a $\gamma\in C_c^{\alpha-1}(\partial\Omega)'$ with $$g=\gamma\circ P\tag1.$$
I was able to show that $P$ is a linear (continuous) surjection and hence, by the fundamental theorem on homomorphisms, $$\hat P:C_c^{\alpha-1}(\mathbb R^d,\mathbb R^d)/\ker P\to\operatorname{im}P=C_c^{\alpha-1}(\partial\Omega)\;,\;\;\;\theta+\ker P\mapsto P(\theta)=\left\langle\left.\theta\right|_{\partial\Omega},\nu_{\partial\Omega}\right\rangle$$ is an isomorphism.
Now let $$\pi:C_c^{\alpha-1}(\mathbb R^d,\mathbb R^d)\to C_c^{\alpha-1}(\mathbb R^d,\mathbb R^d)/\ker P\;,\;\;\;\theta\mapsto\theta+\ker P.$$
We should be able to construct $\gamma$ using $\pi$ and $\hat P$. Maybe we need to apply the fundamental theorem on homomorphisms to $g$, but instead of "dividing" by $\ker g$, we should somehow "divide" by the (closed) subspace $\ker P$ of $\ker g$. But I wasn't able to figure out the details.
At the end we should have obtained a linear map $\gamma$. How can we show that this map is continuous as well?
The context here is superfluous, so let us strip the cumbersome notation. Let $K$ be a field, $P\colon V\rightarrow W$ be a surjective linear map between $K$-vector spaces and $g\colon V\rightarrow K$ be a linear map such that $\ker P\subseteq\ker g$. You are asking why there exists a linear map $\gamma\colon W\rightarrow K$ such that $\gamma\circ P=g$.
Let $\pi\colon V\rightarrow V/\ker P$ be the quotient map. As you note, $P$ factors through the quotient as an isomorphism $\tilde{P}\colon V/\ker P\rightarrow W$ (by the isomorphism theorem). Furthermore, $\ker P\subseteq\ker g$, so $g$ factors through the quotient as a linear map $\tilde{g}\colon V/\ker P\rightarrow K$ (by the universal property of the quotient). Then $\gamma:=\tilde{g}\circ\tilde{P}^{-1}\colon W\rightarrow K$ does the job.