Find the expression for the linear mapping

A real linear space $$W$$ of dimension $$2$$ is considered, with a basis $$B = \{w_1, w_2\}$$. Find the expression $$f(x)$$ of the linear mapping $$f : \mathbb{R^4} \rightarrow W$$ whose kernel is the subspace $$N = {(x_1, x_2, x_3, x_4) : x_1 + 3x_2 + x_4 = 0}$$ and satisfies the condition $$f(0, 0, 0, 1) = w_1 − w_2$$.

What I tried so far was considering the kernel: $$x_1=-3x_2-x_4\Rightarrow (x_1,x_2,x_3,x_4)=(-3x_2-x_4,x_2,0,x_4)$$ which is $$=x_2(-3,1,0,0)+x_4(-1,0,0,1)$$ which then we have $$N=\{(-3,1,0,0), (-1,0,0,1)\}$$ and now I'm guessing we have to consider $$f(0,0,0,1)=w_1-w_2$$ but I don't know how to continue with that

If $$M=span \{(1,3,0,1)\}$$ then $$\mathbb R^{4}$$ is the direct sum of $$N$$ amd $$M$$. We want $$f$$ to be $$0$$ on $$N$$, so we only have to find the value of $$f(1,3,0,1)=c$$, say. Now any point $$(x,y,z,w)$$ can be written as $$a( 1,3,0,1)+v$$ where $$a$$ is chosen such that $$v=(x,y,z,w)- a(1,3,0,1)\in N$$. This means $$x-a+3(y-3a)+w-a=0$$ , so $$a=\frac 1 {11}[x+3y-9+w]$$. We now have $$f((x,y,z,w))=\frac 1 {11}[x+3y+w]c$$. Note that $$f(0,0,0,1)=\frac 1 {11}f(1,3,0,1))$$. So we want $$w_1-w_2=-4f(1,3,0,1))=-4c$$. Finally, $$f((x,y,z,w))=[x+3y+w] (w_1-w_2)].$$