How to simplify $B=\sqrt{3} \tan 70^{\circ}- 4 \sin 70^{\circ}+1$?

Your error: Instead of canceling a factor $4$, that is, equally dividing by $4$ in numerator and denominator, you multiplied by $4$ in the denominator. This is twice the error, first missing to divide and then multiplying, and results in a wrong additional factor $\frac1{4^2}$ for the fraction from then on. And indeed $16\cdot\frac18$ gives the correct result $2$.

Using trigonometric identities and the trig. values at $30°$ one gets in a shorter calculation leaving the denominator unchanged: \begin{align} \frac12B\cos{(70^∘)}&=\cos{(30^∘)}\sin{(70^∘)}+\sin{(30^∘)}\cos{(70^∘)}-2\sin{(70^∘)}\cos{(70^∘)} \\ &=\sin{(100^∘)}-\sin{(140^∘)} \\ &=\sin{(80^∘)}-\sin{(40^∘)} \\ &=2\cos{(60^∘)}\sin{(20^∘)} \\ &=2\cos{(60^∘)}\cos{(70^∘)} \end{align} so that in the end $$ B=4\cos{(60^∘)}=2 $$

Consider the isosceles triangle in the Figure below. Let $\overline{AB} = 2\sqrt 3$ and $\angle CAB = \angle CBA = 70°$. $CH$ is the altitude and $\angle DAB =30°$.

Then you have $\overline{AD} = 2$ and $\overline{DH} = 1$.

Draw from $D$ the line parallel to $AB$ that meets $AC$ in $E$. Also take $F$ on $CE$ so that $\angle FDE = 70°$.

$\triangle ADF$ is isosceles, thus $\overline{DF} = 2$.

$\triangle DEF$ is isosceles so $\overline{EF} = 2$.

$\triangle DFC$ is isosceles so $\overline{FC} = 2$.

$\overline{CE} = 4$, and then $\overline{CD} = 4\sin 70°$.

Your expression comes from the relationship

$$\overline{CH} = \overline{CD}+\overline{DH},$$

that is

$$\sqrt 3 \tan 70° = 4\sin 70°+1.$$

I'll review your steps from here (up to this step everyhing is correct):

\begin{eqnarray} B&=& \frac{2\sqrt 3 \cos (40) + 2 \sin(40)-2\sqrt 3\cos(80)-2\sin(80)}{\sqrt 3 \cos (40)-\sin(40)}=\\ &=& \frac{4\left[\frac{\sqrt 3}2 \cos (40) + \frac12 \sin(40)-\frac{\sqrt 3}2\cos(80)-\frac12\sin(80)\right]}{2 \left[\frac{\sqrt 3}2 \cos (40)-\frac12\sin(40)\right]}=\\ &=&2\frac{\sin (100) -\sin (140)}{\sin(20)}. \end{eqnarray} Then everything is correct again, I think.