If a point $p$ has an i-ward velocity of 1 meter per second and a j-ward velocity of 1 meter per second, then it will move from (0,0) to (1,1) in one second. $$ (\vec i + \vec j )\cdot \delta t $$
If, on the other hand, it has an i-ward velocity of 2 meters per second for HALF a second, then replaced by a j-ward velocity of 2 meters per second for HALF a second, then it will ALSO move from (0,0) to (1,1) in one second. $$ (2\vec i + 2\vec j )\cdot \frac{\delta t}{2} $$
(In the first case, the i and j components act simultaneously, in the second case they act sequentially.)
Given that $p$ goes from (0,0) to (1,1) in unit time in both cases but the paths by which it does so is different (the first path is of length $\sqrt2$ whereas the second is of length $2$), how come we never hear of this second type of linear combination? Taking into account time, the two types of linear combination seem to be conceptually different.
The key insight is that velocity corresponds to displacement while speed corresponds to path length. Your two expressions give displacements, not path lengths:
displacement = $\color\red{\text{average velocity}}$ $\times$ time elapsed
in Scenario 1, displacement magnitude $= \left|\color\red{(\vec i + \vec j)} \times t \right|=t\sqrt2 $
in Scenario 2, displacement magnitude $= \left|\color\red{\frac12(v_1+v_2)}\times t \right|= \left|\color\red{\frac12(2\vec i + 2\vec j )}\times t \right|=t\sqrt2 $
path length = $\color{violet}{\text{average speed}}$ $\times$ time elapsed
in Scenario 1, path length $= \color{violet}{\left|\vec i + \vec j \right|}\times t=t\sqrt2 $
in Scenario 2, path length $= \color{violet}{\frac12\left(s_1 + s_2 \right)}\times t =\color{violet}{\frac12\left(|2\vec i| + |2\vec j| \right)}\times t =2t$
Related: Is arc length displacement or distance traveled?