How can I use the Kirchhoff Laws where there’s a ground connection? Normally we use the Kirchhoff Laws on “loops” in the circuit and to form loops, “electronic components” are connected to wires on both sides. However, with a ground connection there’s only one side connected to the circuit by wire. This makes them seem outlying.
Example:

Find the current through the cells, and the potential difference between each cell.

$\scriptsize\text{Note: I use quotes, “  ”, for unfamiliar physics terminology… Maybe there are mistakes.}$

 A: For KCL, ground is no different than any other node: the sum of the currents into ground must be zero. So, in this case, there is zero current through ground.
For KVL, without ground you get to choose which node is 0V. By convention, ground is 0V, so by connecting ground, you've made the choice.
A: What you call the "ground connection" is simply assigning a node as being zero potential. It is irrelevant to determine the loop current in your diagram since only potential differences and not absolute potentials are relevant.
However, if you were doing Kirchhoff node analysis instead of loop analysis, where the objective is to determine the voltages at different nodes instead of currents in loops, then it is necessary to assign some node as being zero potential.
Hope this helps.
A: Since there's no other connection to ground in the circuit, then this connection to ground has no effect on the loop currents, so it can be ignored for loop or mesh analysis. Its only effect on node voltages is to tell you which node to use as the reference. It doesn't change the potential difference between any two nodes in the circuit.
If there were another ground connection in the circuit, then of course it could cause another loop to be formed, which must be considered for loop or mesh analysis. In this case it is a good idea to draw the connections between ground symbols explicitly to be sure you haven't missed a loop in your analysis.
A: You can usually treat that ground connection as a open circuit (no current), that provides you with an offset for the absolute value of the potential (voltage) in the circuit.
The point connected to the ground and the ground itself has the same potential, here assumed to be $0 \, V$.
Remember that potentials (and thus voltage) are defined up to an additive constant, and the physics is independent on that since the physical quantities are related to the derivatives of the potentials.
