How to use equation of motion like this here? So I will explain my problem exactly what it is.Please do answer in such a format by telling me places where I am wrong from the text I write.I posted this question earlier also but I noticed many of them to give answers which I didn’t even ask and had to explain all the things again in the comments. In the end , some of them leave in between or some don’t even tell the reason why the post is not good.So please check upon these points.
Questions is as follows:
Q :
particle of mass 1 kg has a velocity of 2 m/s. A constant force of 2 N acts on the particle for 1 s in a direction perpendicular to its initial velocity. Find the velocity and displacement of the particle at the end of 1 second.
My questions:

Doubt regarding direction of force:
Force means push or pull.
In image 1,
I drew the force in downwards direction or pulling force (downward vector) which is perpendicular to u (right direction)
Image 2
I drew the force in upwards direction or pushing force (upward vector) which is perpendicular to u (right direction).
So which of these 2 images is right ? Question doesn’t specify it. Also, whether what I though is right or not?
Doubt 2::
In the solution, they have made the equations of motion into two mutually perpendicular vectors.


So I thought of it for a while and thought that maybe because v is velocity, a is acceleration.So they are vectors and have a direction and can be drawn like this.
Q1 I am sure you must have had noticed the direction of vector which is in upward direction.So is it necessary for it to be in upward direction or it can be in downward direction as well.
Q2 why is t also drawn along with those vectors since time is a scalar quantity.
Q3 v = u + at in this equation. They have written v as 2$\sqrt{2}$m/s (Used by solving it in the vectors method )but if you use the equation directly. Then you get v = 2+2(1) = 4m/s.
Why are both of them different.
Q4 why did we consider direction of vector v in a direction i.e in between of u and at. Since force was supposed to there at $” at “$. It looks as if at^2 is equivalent to F = ma in the diagram. Why did that happen ?
Ok.So I am not saying that you exactly answer to me one by one. But in such a way that you cover all my doubts and not that just leave one line and don’t even respond after that.
Do tell me the exact line form the text that you didn’t understand.
 A: 
Force means push or pull.

Doesn't matter.

So which of these 2 images is right ? Question doesn’t specify it. Also, whether what I though is right or not?

It doesn't matter. When the question says velocity they mean the magnitude only in this case (we can see that from the wording). Sure, the direction of the force will determine wether the new velocity vector tilts one way or the other - but in both scenarios the magnitude will be the same. So it doesn't matter from which direction the force pushes, as long as it is perpendicular.

Q1 I am sure you must have had noticed the direction of vector which is in upward direction.So is it necessary for it to be in upward direction or it can be in downward direction as well.

Again, it doesn't matter. The direction doesn't matter at all - that just depends on from where you are looking at the scenario - only the relative direction matters. And that is already fixed with the force being always perpendicular to initial velocity.

Q2 why is t also drawn along with those vectors since time is a scalar quantity.

There is no rule against including scalars in vector equations. If I multiply the scalar $2$ with the vector $(1,3)$ then I get a new vector $2\cdot (1,3)=(2,6)$. No problem doing this.

Q3 v = u + at in this equation. They have written v as $\sqrt 2$m/s (Used by solving it in the vectors method )but if you use the equation directly. Then you get v = 2+2(1) = 4m/s.

No, they have not written $v$ as $\sqrt 2$ m/s. They have written $|v|$ as $\sqrt 2$ m/s. Be careful with what is a vector ($v$) and what is a vector's magnitude ($|v|$).
If you look only in the perpendicular direction, then your equation is correct, though. Just remember to input the values that apply to this perpendicular direction also. The perpendicular acceleration $a$ comes from the force, and that acceleration can be calculated to be $F=ma\Leftrightarrow a=F/m=2\,\mathrm N\,/\,1\,\mathrm{kg}=2\,\mathrm{m/s^2}$. Using this acceleration value with your motion equation in the perpendicular direciton we get:
$$v_\perp = u_{\perp} + at=0+2\,\mathrm{m/s^2}\cdot1\,\mathrm s=2\,\mathrm{m/s}$$
So, here we have the speed component that is added to the motion in the perpendicular direction. The acceleration does not act along with the motion (in the direction of the initial speed) so there is no change along that parallel direction. In that direction the speed stays constant $$v_\parallel=u_\parallel+\underbrace{a}_0t=2\,\mathrm{m/s}+0\cdot 2\,\mathrm{s}=2\,\mathrm{m/s}$$ So, your equation v = 2+2(1) = 4m/s is not correct. With the perpendicular and the parallel components we can now find the magnitude via Pythagora's Theorem:
$$|v|^2=v_\perp^2+v_\parallel^2\quad\Leftrightarrow\quad\\ |v|=\sqrt{v_\perp^2+v_\parallel^2}=\sqrt{(2\,\mathrm{m/s})^2+(2\,\mathrm{m/s})^2}=\sqrt{8\,(\mathrm{m/s})^2}=\sqrt{8}\,\mathrm{m/s}=2\sqrt 2\,\mathrm{m/s}$$

Q4 why did we consider direction of vector v in a direction i.e in between of u and at.

Because it has a direction. The motion started with only a speed component in one direction (the initial direction), $u=(u_\parallel,0)$, but ends out with a speed component in both that and now also in the perpendicular direction, $v=(v_\parallel,v_\perp)$. This is a new vector that has been tilted a bit. It has an angle.

Since force was supposed to there at ”at“. It looks as if at^2 is equivalent to F = ma in the diagram. Why did that happen ?

Not understood. Please clarify this part.
A: Galilean invariance tells us the motions in $x$ and $y$ directions are independent of each other.
In the $x$ direction motion is uniform, so velocity $v_x$ is constant. The displacement is:
$$\Delta x=v_x\Delta t$$
In the $y$ direction there is acceleration, so:
$$v_y=a_y\Delta t$$
The $y$ displacement is:
$$\Delta y=\frac12 a_y(\Delta t)^2$$
With $\text{N2L}$:
$$F_y=ma_y$$
$$a_y=\frac{F_y}{m}$$
Using Pythagoras, total velocity:
$$v^2=v_x^2+v_y^2$$
and total displacement:
$$(\Delta s)^2=(\Delta x)^2+(\Delta y)^2$$
