This first part is an extended comment: I think you have your scaling wrong.
Let $u_k = \phi(\lambda_k(s-s_k), \mu_n(t-t_k))$, where $\phi$ has compact support in the ball of radius 1.
We assume that $(s_k, t_k) \to (1.5,0)$ from within $\tilde{\Omega}$.
That $\mu_k > 2/t_k$, and $\lambda_k > 2$, so $u_k$ has compact support in $\tilde{\Omega}$.
For large $k$, the LHS is approximately
$$ \int |u_k(s,t)|^p s^{m-1} t^{n-1} ~ds ~dt \approx \int |\phi(\lambda_k(s - s_k), \mu_k(t-t_k))|^p t^{n-1} ~ds ~dt $$
Our assumption on $\mu_k$ implies that the support is concentrated near $|t-t_k| < t_k/2$, so we can further replace by
$$ \approx t_k^{n-1} \int |\phi(\lambda_k s, \mu_k t)|^p ~ds ~dt = t_k^{n-1} \lambda_k^{-1} \mu_k^{-1} \|\phi\|_{L^p}^p$$
The RHS on the other hand is
$$ \int |\nabla u_k|^2 s^{m-1} t^{n-1} ~ds ~dt \approx $$
$$ \int \left[ \lambda_k^2 |\partial_s \phi(\lambda_k (s-s_k), \mu_k(t-t_k))|^2 + \mu_k^2 |\partial_t \phi(\lambda_k(s - s_k), \mu_k(t-t_k))|^2 \right] t^{n-1} ~ds ~dt $$
The support property for $t$ gives
$$ \approx t_k^{n-1} (\lambda_k \mu_k^{-1} + \lambda_k^{-1} \mu_k) \|\nabla \phi\|_{L^2}^2 $$
For the desired inequality to hold you need
$$ t_k^{(n-1)(1/2 - 1/p)} (\lambda_k^{1/p+1/2} \mu_k^{1/p-1/2} + \lambda_k^{1/p-1/2} \mu_k^{1/p+1/2} ) \overset{?}{\gtrsim} 1$$
Let $\lambda_k = \kappa \mu_k$ for $\kappa > 1$, this requires
$$ t_k^{(n-1)(1/2 - 1/p)} \mu^{2/p} (\kappa^{1/p+1/2} + \kappa^{1/p-1/2}) \overset{?}{\gtrsim} 1 $$
Choose $\mu_k = 3/t_k$, we need
$$ t_k^{(n-1)(1/2 - 1/p) - 2/p} \overset{?}{\gtrsim} 1 $$
and since $t_k\to 0$ we need
$$ 2/p \geq (n-1)(1/2-1/p) $$
which requires
$$ 1/p \geq \frac{n-1}{2(n+1)} $$
or
$$ p \leq \frac{2(n+1)}{n-1} $$
This makes sense as in the limit where the function is concentrated near one of the axes, you should be looking at a function defined on $\mathbb{R}\times\mathbb{R}^n$, radial in the second variable. So the Sobolev scaling should be computed based on the total dimension $(n+1)$, not on $n$.
Now the proof should be fairly simple.
First, for convenience, use polar coordinates so you have $s = r\cos\theta$ and $t = r\sin\theta$. You are integrating with the measure
$$ r^{m+n-1} ~dr~ \cos(\theta)^{m-1} \sin(\theta)^{n-1} ~d\theta $$
You also have $|\nabla u|^2 = |\partial_r u|^2 + \frac1{r^2} |\partial_\theta u|^2$. Since you are working on the annulus, you have that $r$ is uniformly bounded between $[1,2]$.
On the sector, you further have that $\sin(\theta) \approx \theta$ and $\cos(\theta) \approx $\pi/2 - \theta$. So rescaling once your desired Sobolev inequality is equivalent to
$$ \|u \|_{L^p_w} \lesssim \|\nabla u\|_{L^2_w} $$
on the domain $(z,\theta) \in [0,1]^2$
where the weighted measure is
$$ \theta^{n-1}(1-\theta)^{m-1} ~dz~d\theta $$.
Now fix a cut-off function $\chi(\theta)$ that is identically 0 for $\theta < 1/3$ and identically $1$ for $\theta > 2/3$.
Now,
$$ \|\partial_\theta (\chi u)\|_{L^2_w} \leq \| \chi \partial_\theta u\|_{L^2_w} + \|\chi' u\|_{L^2_w} $$
For the final term you have that since $\chi'$ is supported away from $\theta = 0,1$, you have can apply Poincare's inequality to $u$ which vanishes at the end points, so that
$$ \|\chi' u\|_{L^2_w} \lesssim \| \partial_\theta u\|_{L^2_w}$$
And so with the cut-off being non-harmful, it suffices to prove that
$$ \|\chi u\|_{L^p_w} \leq \|\nabla (\chi u)\|_{L^2_w} $$
and
$$ \|(1-\chi)u\|_{L^p_w} \leq \|\nabla (1-\chi)u\|_{L^2_w} $$
For each individual one, now you can think of $\chi u$ as a function defined on $[0,1] \times B(0,1)$ where the ball is considered as a subset of $\mathbb{R}^n$. And your claim should now follow from the standard Sobolev inequality in dimension $n+1$.
