Why isn't the area under the sine curve from $0$ to $\frac{\pi}{2}$ equal to the area of a quarter of the unit circle?

Imagine a point $P$ moving along the unit circle from $0$ to $\pi/2$. It has a speed in the $x$-direction, and a speed in the $y$-direction (I'll call these $x$-speed and $y$-speed). Note that the $x$-speed of $P$ is constantly changing, and this is reflected in the graph that it draws (the unit circle): we are drawing the $y$-position, but with a changing $x$-speed.

Whereas, the sine graph is drawing the $y$-position with respect to the angle: this is a constant speed. Near an angle of $0$, the point $P$ has a very small $x$-speed, but the sine graph already has a significant $x$-speed. This results in the sine graph covering more area near the start of the movement of $P$ than $P$ does. Thus, the area under the sine wave is bigger than the area under the unit circle.

You could imagine taking the top half of the unit circle and stretching it at the ends where it hits the $x$-axis, to compensate for the slow $x$-speed, and turning it into the sine graph.

The usual integral representation of the area of the quarter-circle is $$\int_0^1 \sqrt{1 - t^2} dt ,$$ where $t$ measures distance along the horizontal axis. By definition, the variable $t$ and the standard angular variable $x$ are related by $t = \cos x$. Substituting this relation into the second integral gives the equivalent integral $$\int_{\pi / 2}^0 \sqrt{1 - \cos^2 x} \cdot -\sin x\, dx = \int_0^{\pi / 2} \sin^2 x \,dx$$ written with respect to the angular variable $x$, but it does not coincide with the original integral. From this point of view, the computation in your question omits the factor of $\sin x$ in the differential equation $dt = -\sin x \,dx$, that is, neglects the Chain Rule.