= Solution
Let $G_0(k),G_l(k)$ denote the <Half-range Fourier transforms> of $g_0,g_l$, and let $N_0(k),N_l(k)$ denote those of the upward <derivatives> $n_0,n_l$. Also set
$$
H(k)=\int_0^le^{ky}h(y)\,dy,\qquad
W(k)=\int_0^le^{ky}v(y)\,dy.
$$
Integrate the tangential <derivatives> in the three spectral <functions> by parts. For example, $\int_0^\infty e^{-ikx}g_0'=-h(0)+ikG_0$, and $\int_0^le^{ky}h'=e^{kl}h(l)-h(0)-kH$. All corner values cancel in $\rho_L+\rho_B+\rho_T=0$. The result is the <semistrip Laplace spectral global relation>
$$
e^{kl}N_l(k)-N_0(k)+k[G_0(k)-e^{kl}G_l(k)]-W(k)-ikH(k)=0.
$$
For $\kappa>0$, use the <sine transforms>
$$
G_j^s(\kappa)=\int_0^\infty\sin(\kappa x)g_j(x)\,dx,\qquad
S_j(\kappa)=\int_0^\infty\sin(\kappa x)n_j(x)\,dx.
$$
Reality of $q$ makes $W(\kappa)$ and $W(-\kappa)$ real, so taking imaginary parts removes the unknown left-side <normal derivative>. At $k=\kappa$ and $k=-\kappa$ the two consequences of the <global relation> are
$$
\begin{aligned}
e^{\kappa l}S_l-S_0
&=\kappa[e^{\kappa l}G_l^s-G_0^s-H(\kappa)],\\
e^{-\kappa l}S_l-S_0
&=\kappa[G_0^s-e^{-\kappa l}G_l^s-H(-\kappa)].
\end{aligned}
$$
Subtracting these equations and then substituting back gives the <semistrip Dirichlet-to-Neumann sine transforms> entirely in terms of the prescribed <Dirichlet boundary data>:
$$
\boxed{S_0(\kappa)=\frac{\kappa}{\sinh(\kappa l)}
\left[-\cosh(\kappa l)G_0^s(\kappa)+G_l^s(\kappa)
+\int_0^l\sinh(\kappa(l-y))h(y)\,dy\right],}
$$
$$
\boxed{S_l(\kappa)=\frac{\kappa}{\sinh(\kappa l)}
\left[-G_0^s(\kappa)+\cosh(\kappa l)G_l^s(\kappa)
-\int_0^l\sinh(\kappa y)h(y)\,dy\right].}
$$
These are the requested transforms of $q_y$ itself. The bottom outward <Neumann boundary data> have the opposite sign. The apparent division at $\kappa=0$ is handled by a continuous limit whenever the corresponding moments exist; there is no singularity for any $\kappa>0$.
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