I have added another picture to further clarify what i am asking - when transmitting, how does the current balun in the drawing below prevent Ishield-outer ?
I'm going to ignore the arrows in your picture and adopt a convention of "positive current sign = upwards", to hopefully avoid confusion. I'm also adding two more measurements, I(balun left side) and I(balun right side) for the currents in each of the two coils.
KCL tells us that I(shield inner) + I(shield outer) - I(balun left side) = 0. [eqn 1]
A quick glance tells us that I(center) = I(balun right side). [eqn 2]
The properties of coax tell us that I(center) = -I(shield inner). [eqn 3]
The balun does its best to ensure that I(balun left side) = -I(balun right side). [eqn 4]
Let's assume that the balun is perfect and that the last equation holds exactly. Then by substituting eqn 4 into eqn 2 we can come up with I(center) = -I(balun left side).
Substituting that into eqn 3 we get I(shield inner) = I(balun left side).
Now substitute that into eqn 1 and we have I(shield outer) + I(shield inner) - I(shield inner) = 0. Or, simplifying, I(shield outer) = 0.
Equations 1 through 3 are another way of stating that "only common-mode current flows on the outside of the shield". If upward current on the outside of the shield makes a "U-turn" and becomes a downward current on the inside of the shield, it induces an upward current on the center conductor in doing so! So perhaps another way of answering your question would be: it does do that, but contrary to your intuition, that doesn't prevent the balun from working — it's exactly what makes the balun work. Any current induced by the outside world on the outside of the shield will be presented to both balun terminals equally, and therefore be subject to the balun's high impedance towards common-mode current.