Consider this situation:
When the box is at the bottom of the frictionless incline, it will have a velocity of $v_f$. The person is an inertial frame of reference that moves at a constant velocity of $v_f$.
From the person's frame of reference, the box has kinetic energy when it is at the top of the ramp. Even if it is moving in the negative direction, velocity would be squared in $KE = mv^2/2$, so now the box has both potiental energy and kinetic energy.
When the box is at the bottom of the incline it is going to have no kinetic energy from the person's frame of reference. How is it that the box had both potential energy ($mgh$) and kinetic energy ($mv^2/2$) at the top of the frictionless incline but had neither potential nor kinetic energy at the bottom?
$$mgh + \frac{mv^2}{2} = 0$$
The law of conservation of energy says energy is transferred, but not lost. Where has the energy gone?
Basically the law of conservation of energy should not be violated no matter the frame of reference, but the final formula does that. Where has the energy gone?
Best Answer
Consider two reference frames $S$ and $S'$. Assume that $S'$ moves with a velocity $\mathbf{u}=u\mathbf{\hat{x}}$ relative to $S$ such that the origins of $S$ and $S'$ coincide at $t=0$ and their axes remain parallel. Further assume that the surface of the inclined plane is stationary in $S$ as shown in the figure (the dashed curve can be ignored till the end of the discussion).
There are two points to consider before proceeding further.
The fundamental relation for energy considerations is the work-energy theorem which states that the change in the kinetic energy of a particle as it moves from an initial point $A$ to a final point $B$ is equal to the work done $$K_B - K_A = \int_{t_A}^{t_B} \mathbf{F}(t) \cdot \mathbf{v}(t) \, dt$$ where the force $\mathbf{F}(t)$ can arise from a constraint, be time-dependent or be non-conservative. The work-energy theorem is frame-independent. It is only in the special case of time-independent and conservative forces that one can identify a scalar potential energy $U$ such that $\mathbf{F} = - \nabla U$. Then the work energy theorem boils down to $K+U=\mathrm{constant}$.
The block is confined to move on the surface of the inclined plane of constant inclination $\theta$. In the frame $S$, this static (scleronomic) constraint can be expressed as $$x \, \tan\theta + y - h=0.$$ In the frame $S'$, the surface of the inclined plane is non-stationary and this moving (rheonomic) constraint can be written as $$x \, \tan\theta + y - H(t)=0$$ where $H(t)=h - u \tan\theta \,\, t$. Forces arising from scleronomic constraints do no work since they are orthogonal to the velocity. However forces arising from rheonomic constraints can perform real work since the net velocity of the particle can have a component along the direction of the constraint forces. See section 2.1 of Jose and Saletan for a beautiful discussion of this concept.
Newton's law in either frame is $$m \dot{\mathbf{v}} = -mg {\hat{\mathbf{y}}} + \mathbf{F}_c(t)$$ where $\mathbf{F}_c$ is the force of constraint. As done in proving the work-energy theorem, we multiply by $\mathbf{v}$ and integrate with respect to time to get $$K_B - K_A = mg \big[ y(t_A) - y(t_B) \big] + W_c$$ where the work done by the constraint forces is $$W_c = \int_{t_A}^{t_B} \mathbf{F}_c(t) \cdot \mathbf{v}(t) \, dt \qquad \textrm{in }S$$ and $$W_c' = \int_{t_A}^{t_B} \mathbf{F}_c'(t) \cdot \mathbf{v}'(t) \, dt \qquad \textrm{in }S'.$$ It is easily shown that $$\mathbf{F}_c(t) = \mathbf{F}_c'(t) = mg \cos\theta \, (\sin\theta \mathbf{\hat{x}} + \cos\theta \mathbf{\hat{y}} ).$$ Also it is easy to show that the velocity $\mathbf{v}$ and the position $\mathbf{r}$ of the block in $S$ are $$\mathbf{v} = g t \sin \theta \, (\cos\theta \mathbf{\hat{x}} - \sin\theta \mathbf{\hat{y}} )$$ $$\mathbf{r} = \frac{g t^2 \sin \theta \cos\theta}{2} \mathbf{\hat{x}} + \Big( h - \frac{g t^2 \sin^2 \theta}{2} \Big) \mathbf{\hat{y}}.$$ The corresponding quantities in $S'$ are obtained via the Galilean transformation $$\mathbf{v}' = \mathbf{v} - \mathbf{u}$$ $$\mathbf{r}' = \mathbf{r} - \mathbf{u} t.$$ We find $\mathbf{F}_c \cdot \mathbf{v} = 0$ and hence $W_c=0$ while $$W_c'= -\int_{t_A}^{t_B} \mathbf{F}_c(t) \cdot \mathbf{u} \, dt = -mug \sin\theta\cos\theta (t_B-t_A).$$
Therefore the work-energy theorem in $S$ is $$K_B - K_A = mg \big[ y(t_A) - y(t_B) \big]$$ while in $S'$ it takes the form $$K_B' - K_A' = mg \big[ y'(t_A) - y'(t_B) \big] - mug \sin\theta\cos\theta (t_B-t_A).$$
We can now finally come to the particular problem asked by the OP. Take the points $A$ and $B$ to be respectively those points where the inclined surface meets the $y$ and $x$ axes. The time $T$ taken by the block to slide down the incline from a height $h$ is obtained from solving $y(T) = 0$ as $$T = \frac{\sqrt{2h/g}}{\sin\theta}$$ and at this instant of time $\mathbf{v}(T) = \sqrt{2gh} \, (\cos\theta \mathbf{\hat{x}} - \sin\theta \mathbf{\hat{y}} ).$
We now verify the work-energy theorem
in frame $S$ $$K_B-K_A=\frac{m}{2} \big[ v(T)^2 - v(0)^2] = mgh$$ while $$mg \big[ y(t_A)-y(t_B) \big] = mgh$$
in frame $S'$ $$K_B'-K_A'=\frac{m}{2} \big[ v'(T)^2 - v'(0)^2] = \frac{mg^2T^2}{2} \sin^2\theta - mugT \sin\theta\cos\theta$$ while $$ mg \big[ y'(t_A) - y'(t_B) \big] - mug \sin\theta\cos\theta (t_B-t_A) = mgh-mugT \sin\theta\cos\theta $$ which, using the relation between $h$ and $T$, is the same as $K_B'-K_A'$.
Thus the work-energy theorem is verified in both frames.
The OP asked for the particular case of $u=v_x(T)=\sqrt{2gh}\cos\theta=gT\sin\theta\cos\theta$ and considered energy conservation in $S'$. In this case - $$K_B'-K_A'= \frac{mu^2}{2} (\sec^2\theta-2)$$ while $$ mg \big[ y'(t_A) - y'(t_B) \big] - mug \sin\theta\cos\theta (t_B-t_A) =\frac{mu^2}{2} (\sec^2\theta-2) $$ Notice that $K_B' \neq 0$ -- this is because the $v_y(T)\neq 0$ as the OP implicitly assumed. The reason is that the OP was looking at the case where no motion in the vertical direction would be allowed when $y=0$, i.e., for $t>T$. However this would require that the constraint force (normal reaction) change discontinuously. A better solution would be to consider the block sliding down a smooth curve (such as the dashed curve in the figure). In this case $v_y(T)=0$ and the reaction force would also vary smoothly. The work-energy theorem would, ofcourse, still be valid.
NOTE:
The point that moving constraints can do real work was mentioned in the answer by Pygmalion.
This problem of a block sliding on a moving inclined plane is considered in section 3.9 of Strauch.
Worrying about the mass of the wedge or of the earth is misleading. We can just consider a particle constrained to move on a mathematical surface subject to a uniform and constant body force in the $y$-direction. This surface could also be moving. For this particular case, think of the beads sliding down a tilted abacus made of thin massless wires. You look at this situation from a frame at rest w.r.t. the abacus and then in a moving frame.