Why does a longer observation time improve DFT resolution, but repeating a signal does not?
Why is this not equivalent to simply observing the signal for 1 period, and then paste it together N times?

It's only equivalent if certain conditions are met. Let's look at a single sine wave with frequency $f_0$ and period $T_0 = 1/f_0$. Sample rate is $f_s$, sampling period $T_s = 1/f_s$ and DFT length is $N$.
The input to the FFT would be
$$x[n] = sin(2\pi \cdot f_0\cdot T_s \cdot n), 0 \leq n &lt; N$$
Now let's create two signals of length $2N$: one with double the length and one with pasting two buffers together.
$$x_1[n] = sin(2\pi \cdot f_0\cdot T_s \cdot n), 0 \leq n &lt; 2N$$
and
$$x_1[n] = sin(2\pi \cdot f_0\cdot T_s \cdot mod(n,N)), 0 \leq n &lt; 2N$$
The only difference is that modulo function in the second one, but it's important. It basically resets the counter to 0 when it reaches an integer multiple of N. In general this will create a discontinuity in the signal. It forces $x_2[N] = x_2[0] = 0$ even if $x[N]$ would be non zero. The first halves of $x_1$  and $x_2$ are the same but the second halves are not. $x_1$ is still a sine wave but $x_2$ isn't anymore.
The ONLY case where pasting is the same as making the signal longer is if the period of the signal is an integer multiple of the sampling period. In this case, there is no discontinuity and both cases are identical. And sure enough, in this case the DFT will resolve two different frequencies perfectly (as long as both meet the integer multiple condition).
A more philosophical interpretation: Making the signal longer provides more information. Pasting buffers just repeats information you already have, so you are not gaining anything new.