Electronic – How to properly scale, time-reverse and shift a signal

If I understand your question correctly, it is important to understand that the device works in a couple separate stages.

1. The parallel inputs. These have latches that must be signaled to read the inputs and internally hold their state. Once latched, the signal on the input pins can change but the internal signals will be stable. (RCK)
2. The next step is where these internal signals are copied into the shift register. (\$\overline{\small \text{SLOAD}}\$)
3. The last step is shifting out the bits in the shift register (SCK)

If I understand the truth table and timing diagram in the datasheet correctly then the correct order of steps is the following:

1. Initialize the signals as follows RCK = low; \$\overline{\small \text{SLOAD}}\$ = high; \$\overline{\small \text{OE}}\$ = low; SCK = low;
2. Apply parallel signals to the inputs A~H
3. Clock the parallel signals into the latches by driving RCK high. RCK triggers on a rising edge. Drive RCK low, its inactive state.
4. Copy the contents of the input latches into the shift register by driving \$\overline{\small \text{SLOAD}}\$ low. \$\overline{\small \text{SLOAD}}\$ is active low. Drive \$\overline{\small \text{SLOAD}}\$ high again, its inactive state.
5. Read pin Q_H on your MCU
6. Apply a clock pulse on SCK to shift the data out on pin Q_H. SCK triggers on a rising edge. Drive it high, then low.
7. Repeat last 2 steps for as many inputs as you are interested in.

The system is linear, causal, stable, but not time-invariant.

It is linear because from the definition if \$y_1(t)\$ is the response to input \$x_1(t)\$, and \$y_2(t)\$ is the response to input \$x_2(t)\$, then the response to the input signal \$ax_1(t)+bx_2(t)\$ is given by

$$y(t)=ay_1(t)+by_2(t)$$

It is causal because in order to compute the current output signal, not future values of the input signal are necessary, i.e. to compute \$y(t_0)\$ we only need to know \$x(t_0)\$ (and not even its past values, i.e. the system has no memory).

The system is stable in the BIBO (bounded-input bounded-output sense), because any bounded input signal \$|x(t)|<K\$ produces a bounded output signal \$|y(t)|<L\$ (in our case \$K=L\$).

The system is time-variant because the response to a shifted version of the input signal is not equal to the shifted output signal, i.e. if \$y(t)\$ is the response to \$x(t)\$, then \$y(t-t_0)\$ is generally not the response to \$x(t-t_0)\$. You can see this by noting that the output is always zero for \$-1<t<1\$, no matter how the input signal is shifted. If the system were time-invariant, then also the zero portion of the output would need to be shifted with the input signal.