It is there … I just stopped drawing it in after Time 0 and Time 1.All of these are missing the downward gravity acceleration. You need to add the 1 unit down to every step.
Look at Time 1. The RED vertical component of the thrust vector is 1.996 G V (V for Vertical). Since gravity is -1G V (vertical) … shown in black as 1G … the resulting Vertical Acceleration Vector (in BLUE) is 1.996 G - 1G = 0.996 G V (for Vertical). LOOK, the BLUE arrow is labeled 0.996 G Vertical!
Now look at Time 2. Although the Gravity Vector is not drawn (since it did not change and just cluttered the diagram), but the RED 1.878 G vertical thrust vector - the constant 1G gravity vector yields the 0.878 G vertical acceleration vector in BLUE.
Then when you add the vertical vectors (+ 1.878G - 1G) and horizontal vectors (+0.679G) you get a total acceleration vector for the rocket at Time 2 of 1.112G.
I drew it that way because the drawing represented the math that was done earlier. The math was easiest if one resolves the horizontal and vertical separately. If we really want to focus on strictly GRAPHIC solutions, then draw a vector 2 “units” (G) long at a 45 degree angle and a second vector from the tip down 1 “unit” (G) for gravity. What is the length of the resulting vector connecting the start and finish positions? (Longer than 1!)