Why do two waves moving in opposite directions make a standing wave that doesn't go anywhere?

The pattern stays put because the two waves are still there and still moving. At every point the rope's displacement is the sum of the two travelling displacements; at some points one wave is always the negative of the other, so those points never move, and between them the sum swells and shrinks in place.

Take a rope carrying two waves of equal size, one going right and one going left. Let the wavelength be $\lambda = 2$ metres and the frequency be $f = 1$ cycle per second. Then the wavenumber is $k = 2\pi/\lambda = \pi$ radians per metre and the angular frequency is $\omega = 2\pi f = 2\pi$ radians per second. The speed of each wave is $v = f\lambda = (1)(2) = 2$ metres per second, one to the right and one to the left.

The first term on the left is a wave travelling right, the second a wave travelling left. The right-hand side is a fixed shape in $x$, namely $\sin(kx)$, multiplied by a number that depends only on time, $\cos(\omega t)$. That is what a standing wave is: every point shares the same time dependence, and the places where $\sin(kx)=0$ are zero at every instant. At $x=0$ the right-hand side is $2\sin(0)\cos(\omega t)=0$ for all $t$, so $x=0$ is a node. At $x=0.5$ metres, $\sin(\pi\times 0.5)=1$, so the displacement there is $2\cos(2\pi t)$. At $t=0$ it is $2$; at $t=0.25$ seconds, $\cos(\pi/2)=0$, so it is $0$; at $t=0.5$ seconds, $\cos(\pi)=-1$, so it is $-2$. That point swells from $2$ to $0$ to $-2$, but it never travels along the rope.

The arithmetic shows where the stillness comes from. At a node the right-going wave has displacement $\sin(kx-\omega t)$ and the left-going wave has $\sin(kx+\omega t)$. Their sum is zero because the two sines are equal and opposite at that $x$ for every $t$. At an antinode the two sines are equal, so they add to twice one wave. The standing pattern is not a wave that has lost its motion; it is two waves whose motions are opposite at each point, so the sum has no direction along the rope.

The nodes make it look as if the motion has died, and that is the natural guess that fails. The individual waves have not stopped. With the numbers above, the right-going wave has a crest at $x=0.5$ metres when $t=0$. At $t=0.25$ seconds that crest is at $x=1.0$ metres, because $\pi x - \pi/2 = \pi/2$ gives $x=1.0$. It moved $0.5$ metres in $0.25$ seconds, a speed of $0.5/0.25=2$ metres per second. The left-going crest moved from $x=0.5$ to $x=0.0$ in the same time, also $2$ metres per second but the other way. The two crests pass through each other. What stays put is the sum's pattern, not either travelling wave.

The standing result needs equal amplitudes, the same frequency, and exactly opposite directions. If the amplitudes differ, say $A$ going right and $B$ going left, the sum is $A\sin(kx-\omega t)+B\sin(kx+\omega t)=(A+B)\sin(kx)\cos(\omega t)+(B-A)\cos(kx)\sin(\omega t)$. The second term is not zero when $A\neq B$, so a travelling component remains and the zero points no longer stay at zero. If the frequencies differ, the two waves drift in and out of step, and the pattern moves or beats rather than standing still. If the directions are not exactly opposite, there is a net direction of travel. The fixed nodes belong to the equal, opposite, same-rate case, and to the sum rather than to either wave.