= Solution
Use the printed coordinate derivative $u_x(0,t)$, not an outward <normal derivative>; at the left endpoint these have opposite signs. Assume the usual bounded or admissibly growing solution at infinity, with an initial trace $b_0=u_0(0)$ and forcing possessing a <Laplace transform>. Set $a=\alpha/2$, so $\beta=a^2$, and use $z=\sqrt p$ on the <principal square root> branch. Define
$$
I(z)=\int_0^\infty e^{-zy}u_0(y)\,dy,\qquad F(p)=\int_0^\infty e^{-pt}f(t)\,dt.
$$
The half-line <Dirichlet Green function> for $p-\partial_x^2$ gives
$$
U_D(x,p)=\int_0^\infty\frac{e^{-z|x-y|}-e^{-z(x+y)}}{2z}u_0(y)\,dy,\qquad U_D(0,p)=0,\qquad U_{D,x}(0,p)=I(z).
$$
Write the transformed solution as $U=U_D+e^{-zx}B(p)$. The <Laplace transform of a derivative> in the dynamic <boundary condition> is $pB-b_0$, so
$$
(p+\beta)B-b_0+\alpha[I(z)-zB]=F(p).
$$
The special choice of $\beta$ gives the perfect square
$$
p-\alpha\sqrt p+\beta=(z-a)^2,\qquad \boxed{B(p)=\frac{F(p)+b_0-\alpha I(\sqrt p)}{(\sqrt p-a)^2}.}
$$
In particular, the initial boundary trace cannot be omitted. With
$$
K_D(x,y,t)=\frac{e^{-(x-y)^2/(4t)}-e^{-(x+y)^2/(4t)}}{\sqrt{4\pi t}},
$$
an <integral representation> is
$$
\boxed{u(x,t)=\int_0^\infty K_D(x,y,t)u_0(y)\,dy+\frac1{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}e^{pt-x\sqrt p}\frac{F(p)+b_0-\alpha I(\sqrt p)}{(\sqrt p-a)^2}\,dp.}
$$
Choose the <Bromwich contour> to the right of the forcing's growth abscissa and every pole; $\gamma>\max(0,a^2)$ and sufficiently large for the data is a safe choice for real $a$. No sign of $\alpha$ is explicitly imposed in this question.
The inverse transform can also be performed explicitly. For the <repeated-root dynamic-boundary heat kernel>, set
$$
J_a(x,t)=\mathcal L^{-1}\!\left\{\frac{e^{-x\sqrt p}}{(\sqrt p-a)^2}\right\}(t).
$$
First integrate the <Heat Poisson kernel> against $e^{ar}$, or differentiate the result with respect to $a$:
$$
\mathcal L^{-1}\!\left\{\frac{e^{-x\sqrt p}}{\sqrt p-a}\right\}=\frac{e^{-x^2/(4t)}}{\sqrt{\pi t}}+aE_a(x,t),\qquad E_a=e^{-ax+a^2t}\operatorname{erfc}\!\left(\frac{x}{2\sqrt t}-a\sqrt t\right).
$$
Since differentiation of $(\sqrt p-a)^{-1}$ with respect to $a$ produces $(\sqrt p-a)^{-2}$, and $\partial_aE_a=(-x+2at)E_a+2\sqrt{t/\pi}\,e^{-x^2/(4t)}$, the desired kernel is
$$
\boxed{J_a(x,t)=(1-ax+2a^2t)e^{-ax+a^2t}\operatorname{erfc}\!\left(\frac{x}{2\sqrt t}-a\sqrt t\right)+2a\sqrt{\frac t\pi}\,e^{-x^2/(4t)}.}
$$
Here $\operatorname{erfc}$ is the <complementary error function>. A direct integral characterization, also proving the sign and the transform, is
$$
J_a(x,t)=\int_0^\infty r e^{ar}\frac{x+r}{2\sqrt\pi\,t^{3/2}}e^{-(x+r)^2/(4t)}\,dr.
$$
The <convolution theorem for Laplace transforms> now gives the fully real time-domain representation
$$
\boxed{u(x,t)=\int_0^\infty\left[K_D(x,y,t)-\alpha J_a(x+y,t)\right]u_0(y)\,dy+b_0J_a(x,t)+\int_0^tJ_a(x,t-s)f(s)\,ds.}
$$
All quantities here are known from the prescribed data. For $x>0$ it tends to $u_0(x)$ as $t\downarrow0$. At $x=0$, $J_a(0,t)\to1$ and the spatial correction tends to zero, recovering $b_0$. A solution classical through the initial corner additionally needs $u_0''(0)+\alpha u_0'(0)+a^2b_0=f(0)$; weaker corner regularity does not invalidate the formula for positive times.
The <repeated-root unstable heat boundary mode> gives a useful sign check on the <dynamic boundary condition for the heat equation>. If $a>0$, $(\sqrt p-a)^2$ has a genuine double pole at $p=a^2$ on the physical branch, and the exact homogeneous mode $e^{-ax+a^2t}$ satisfies both the <heat equation> and the printed boundary condition. Generic data can also excite a $t e^{a^2t}$ contribution. A contour deformation must retain this repeated-pole contribution. If $a<0$, the putative root $\sqrt p=a$ is outside the chosen branch and is not a physical pole. At $a=0$, $J_0=\operatorname{erfc}(x/(2\sqrt t))$, as expected when the boundary trace satisfies $b'(t)=f(t)$.
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