Two batteries charging one capacitor

What I am thinking to myself right now: is that I want to do the current law where In = Iout, however, I only have one current going into a node thus I can't find the rest as shown

You're already off track at this point.

You don't need to solve KCL to understand the circuit. You basically already have it solved. The input voltage is +5 V, and the op-amp inverting input is virtual ground. Therefore 2.5 mA through R1, therefore 2.5 mA through R2. Therefore the capacitor node is at -5 V. That's it.

You know to get to -5 V, therefore 0.5 uC must have at some point flowed out of the capacitor to charge it to that voltage.

But wait, the diode has its anode at the op-amp output and its cathode at the capacitor. So the op-amp couldn't have drawn charge off the capacitor this way. You should have been modeling the diode as an open and not a short for this part of the input cycle.

So now go back and analyze this part of the cycle with the diode as an open. And remember that when the negative feedback loop is broken, the inverting input no longer acts as a virtual ground.

^{simulate this circuit – Schematic created using CircuitLab}

The schematic shows how the whole circuit should/could be interpreted. Of course, I have assumed that the second opamp is used as a non-inv. amplifier (gain A). Only in this case, the circuit can be used as a linear amplifier.

Regarding the output Vout1 it can be treated as a classical inverting configuration. For gain calculation you can set Vn to zero (Vn is the voltage at the inv. terminal). Because this voltage is produced from two voltage sources (V1 and Vout1) you have to use superposition rules Vn=Vn1+Vn2 (Vn1 caused by V1 with Vout1=0 and Vn2 caused by Vout1 with V1=0)

If the gain Vout1/V1 is found, it will not be a problem to find the wanted ratio Vout2/Vin because of Vout2=Vout1*A with A=(1+R5/R4).

EDIT/Comment:

Realizing that R3 enters the feedback path with a value reduced by the factor A (because the current through R3 is A times larger if compared with R2), you immediately can write down the result (making use of the well-known inverter formula):

G1=Vout1/V1=-(R2||R3eff)/R1 with R3eff=R3/A

G2=Vout2/V1=A*G1