Apparently the discussion does not go much ahead. So I will try to speed it up and show how it goes. First try to establish the half-reaction with ions, not with molecules and formula units. We will go back to neutral compounds later on and get rid of the positive and negative charges at the end.
Here, in the first half-reaction, the ion $\ce{MnO4-}$ is reduced to $\ce{Mn^{2+}}$, and the $4$ lost oxygen atoms are transformed into $\ce{H2O}$ molecules, here $\ce{4H2O}$ to be added on the other side of the arrow. These $4$ water molecules introduce unwanted $\ce{H}$ atoms, that must appear on the other side of the arrow as $\ce{8 H+}$ ions In the second half-reaction, the ion $\ce{Cl-}$ is oxidized to $\ce{Cl2}$. Hopefully you know how to build these half-equations. If you don't, tell us, and we will explain it in more details. Here are these half-equations :
$$\ce{MnO4- + 8 H+ + 5 e- -> Mn^{2+} + 4 H2O} \\ \ce{2 Cl- -> Cl2 + 2 e-}$$ Now you have to add these two half-equations after multiplication of both in order to make the electrons disappear. This can be done by multiplying the first half-equation by $2$ and the second by $5$. Look. $$\ce{2 MnO4- + 16 H+ + 10 e- -> 2 Mn^{2+} + 8 H2O} \\ \ce{10 Cl- -> 5 Cl2 + 10 e-}$$
This produces $10$ electrons on both sides, so that the electrons can be skipped after summing. $$\ce{2 MnO4- + 10 Cl- + 16 H+ -> 2 Mn^{2+} + 5 Cl2 + 8 H2O}$$ The final equation is obtained. But now you would like to have the ions disappear, so that only neutral species appear in the equation. So you must add $2+10 = 12$ ions $\ce{K+}$ ions on the left hand side to compensate the anions and form neutral potassium salts. You also must add $8$ ions $\ce{SO4^{2-}}$ to compensate the $\ce{H+}$ ions. Afterwards you must take care of these new ions and make some neutral salts on the right hand side. So the two $\ce{Mn^{2+}}$ ions recombine with $2$ of the $\ce{8 SO4^{2-}}$ ions, forming $\ce{8 MnSO4}$. The remaining ions produce $\ce{6 K2SO4}$ . This gives : $$\ce{2 KMnO4 + 10 KCl + 8 H2SO4 -> 2 MnSO4 + 5 Cl2 + 6 K2SO4 + 8 H2O}$$ Is it OK ?