how do i derive this equation v^2(final) - v^2(initial) = 2a(x(final) - x(initial)) from...?

You need to eliminate t so find t in terms of V and a

you have ... Vf = a.t+ Vo

rearrange so that...

t = [ Vf - Vo ] / a

now substitute into the top equation, all instances of t

Xf = (1/2).a.t^2 + Vo.t + Xo

= [Vf - Vo]^2/[2.a] + [Vo.Vf - Vo^2]/a + Xo

Bring Xo across and make 2a the common denominator on Right Hand Side

Xf - Xo = 1/[2.a] * [Vf^2 + Vo^2 - 2.Vf.Vo + 2.Vo.Vf - 2.Vo^2]

Cancel terms in Right Hand Side and multiply both sides by 2a and you're done

2.a[Xf - Xo] = Vf^2 - Vo^2

This one is a little tricky.

Most sane people would start with

v_average=(v_initial +v_final)/2

x=v_average t=((v_i +v_f)/2) t

(v_i +v_f)/2=x/t

and then use

v_f=v_i +at

so

v_f-v_i=at

Multiply the equations together,

(v_f -v_i)(v_f +v_i)/2=(at)(x/t)

Notice the difference of two squares,

(v_f)^2-(v_i)^2=2 ax

And there you have it.

But since you are asked to start with those two kinematic equations, here is that process:

Take

v_f=v_i+at

solve for t

t=(v_f-v_i)/a

plug this in for t in your other equation

x_f=x_i +v_i t+(1/2)at^2

Simplify quite a bit and you get your answer. It took me about 5 lines of simplification to get it to look pretty.