How does the output jump up to 7.5 volts when the switch is closed and opened again?
The successive steps shown in the schematic below will aid in understanding the following discussion points:







simulate this circuit &ndash; Schematic created using CircuitLab


Here, capacitor \$C_1\$ has long been discharged so the voltage difference across its leads is \$0\:\text{V}\$. So there is no current in \$R_1\$ or \$R_2\$ and therefore no voltage drop across either resistor. The output voltage will be \$5\:\text{V}\$.
Now the switch is closed. \$R_2\$ has its own current but doesn't participate in charging \$C_1\$ so I removed it as "distracting" so that you can focus on the capacitor charging activity. \$V_\text{OUT}\$ will rise, exponentially, until it reaches \$V_\text{OUT}=+5\:\text{V}\$ and then the charging process stops.
The switch is re-opened and the circuit returns to the configuration shown in step 1 above. However, there is a difference. Now, node \$N_2\$ is \$+5\:\text{V}\$ relative to node \$N_1\$. Because of this, current immediately starts flowing through \$R_1\$ and then through \$R_2\$ in order to start discharging \$C_1\$ via this path. The two resistors have the same value, so their mid-point (which is attached to a \$+5\:\text{V}\$ voltage source) will be half-way between or \$2.5\:\text{V}\$ lower than node \$N_2\$ and \$2.5\:\text{V}\$ higher than node \$N_1\$. So it must be the case, right after opening the switch, that \$V_{\text{N}_1}=+2.5\:\text{V}\$ and \$V_{\text{N}_2}=+7.5\:\text{V}\$. Note that there is still just the original \$5\:\text{V}\$ voltage difference across \$C_1\$. That didn't change (yet.) But now \$V_\text{OUT}\$ must initially start higher than \$+5\:\text{V}\$ (by the voltage drop across \$R_1\$.) Now, \$C_1\$ discharges via the two resistors and eventually returns to its discharged state. In the meantime, the current through the two resistors declines and therefore also the voltage at \$V_\text{OUT}\$ gradually returns to \$+5\:\text{V}\$.


You can adjust the ratio of \$R_1\$ and \$R_2\$ to get different peak voltages. For example, if you change \$R_1\$ to \$2\:\text{k}\Omega\$ then \$V_\text{OUT}\$ will rise above \$8.3\:\text{V}\$ for a short time -- more than the \$7.5\:\text{V}\$ you see in your trace.