I think you have a problem with Kirchoff's laws.Now it states that the potential drop across a closed loop is 0.
Your txt seems to have taken the direction of I as the direction of electron flow.Now suppose you move along the wire anticlockwise(The txt has moved clockwise.) Across the resistor ,potential drops but you travel along the original direction of current.So the potential drop becomes positive so $iR$.Across the capacitor you move from the positive to negative plate.Obviously the potential drop .So the potential drop along this direction is positive so $+Q/C$.
For what your txt says ,they move clockwise.Along the resistor, as you move opposite to the direction of current the potential rises along the direction you move.So the potential drop is negative.Similar is the case for the capacitor.
Note:May be you are confused by the direction of current,but thats just a convention.Your book seems to have used another convention.
2.Now coming to the video of Youtube.I suppose you have read the first part of the answer as it explains the equations and contains the most important part potential drop.
Well in the video it says,$V_c+V_r=0$.Now think one thing-What is $V_c$-it is the potential difference!So is $V_r$.What are they- positive right?So it says positive +positive=0!!!!kidding...
Actually it should be $V_c=V_r$.Now their blunder lies here...
What you should do- just forget these potential differences and stuff and their equating.You have Kirchoff's laws for the same thing.The laws say that Potential dropis zero-why?As the electrostatic field is conservative-so the poential drop (or rise) is 0.But the blunder the videomaker makes is he takes both potential rise and drop together...
I hope now you understand the importance of potential rise or drop.Yes that makes it mechanical,artificial but what to do?Suppose you have a circuit like this-
In the right loop hoe would you do it by physicsl sense?You have to apply the laws blindly.Okay even you pull it off here there exists more complex circuits.
Okay,now after this as they made this blunder it leads to their next blunderOut of the middle of nowhere the introduce a negative sign-why?to make the current positive!!!But maths equations in physics have the fame of giving you the correct signs at last.So why manipulate signs midway?
In the last part even if you do the correct way using potential drop,you will get i as negative.That is because we have and your text book has chosen the wrong direction of current.You might be wondering why i say so-i just said its different convention.BUT in the writing of equations somewhere a fault has crept in.Wanna know where??It is in the potential drop of the capacitor.
You took the potential to decrease from the positive plate to the negative plate.But that happens only whan there is positive charges moving.
Aha..So the equations speak.They are saying -look man you chose the current to be clockwise...But my $I$ comes as negative So its in the opposite direction
Thats the significance of the minus.But IF you manipulate the signs midway like the video did you are actually making the equations dumb!!!It is now you controlling the direction of current.But you dont want to do that do you-there will be places where you will be helpless only equations and signs will assist you in the direction of current.
But unfortunately the video seems to care about none of these.They just exploited the simplicity of the situation using DEAD WRONG and MISLEADING methods.
So i hope you understand my points and the mistakes of the video.If i can think of some more counterarguments i will update it.