Yet another view of the tangent surface:
 
    Please recall that the curve is given parametrically by:

• The portion of tangent line that is shown is between the planes  x = -1  and x = 1.
     {Thus, the point of tangency is not included for any t-value with  t > 1  or  t < -1.}
   
• Each segment has one endpoint on the plane  x = -1  and/or one endpoint on the plane  x = 1.
     {Some segments have been "trimmed", to make their length ≤ 2.}

Portions of the following curves are traced in the figure:

• The twisted cubic, shown in dark blue.
• The intersections of the tangent surface with the planes  x = 1  and  x = -1, 
  shown in blue-green and dark green respectively. Each of these curves has a cusp
  at the point where it meets the twisted cubic.
   

The implicit equation of the tangent surface.
    By substituting from the parametrization, we obtain:

The osculating plane at a point of the twisted cubic,
           and the tangent cone (of the tangent surface) at the same point.
For simplicity, let's consider the situation at the origin. The lowest degree homogeneous component of the defining equation of the surface is  z²,  so the tangent cone at the origin is given by the equation  z² = 0.  On the other hand, the parametrization of the curve is given by the vector valued function  f(t) = (t,t²,t³),  so that  f'(0) = (1,0,0)  is a tangent vector at the origin. Furthermore, the osculating plane to the curve at the origin corresponds to the vector space spanned by  f'(0)  and  f"(0) = (0,2,0).  {In general, the principal normal to the curve at the origin is perpendicular to  f'(0)  and lies in the subspace of  R³  spanned by  f'(0)  and  f"(0) = (0,2,0).  The fact that  f"(0)  happens to be perpendicular to  f'(0)  is specific to the parameter value  t = 0.}  Anyway, the osculating plane at the origin is the plane  z = 0.  This means that the tangent cone to the surface at the origin coincides with the osculating plane to the curve at the origin --  or more precisely, the tangent cone is equal to the osculating plane, counted with multiplicity = 2.  
    In the case of the twisted cubic, one can check directly that at every point of the curve, the tangent cone to the tangent surface is equal to the osculating plane of the curve at the same point  --  counted with multiplicity 2 of course. The verification is somewhat intricate and is not presented here.  I suspect that this is a particular instance of a theorem about space curves, but I don't know a reference for such a theorem.