The average velocity of a particle during some elapsed time $\Delta t$ is, in words, the constant velocity that gives the same displacement in the same elapsed time.
Mathematically, the average velocity is given by
$$\mathbf{v}_{avg} = \frac{\Delta \mathbf{r}}{\Delta t}$$
where $\Delta \mathbf{r} = \mathbf{r}_f - \mathbf{r}_i$ is the displacement vector and $\Delta t = t_f - t_i$ is the elapsed time during which the displacement took place.
For example, consider the case that a particle moves with constant velocity $1 \mathrm{\frac{m}{s}} \hat{\mathbf{x}}$ for 4 seconds and then with constant velocity $1 \mathrm{\frac{m}{s}} \hat{\mathbf{y}}$ for 3 seconds.
The displacement vector for the 7 seconds of motion is, by inspection,
$$\Delta \mathbf{r} = (4\hat{\mathbf{x}} + 3 \hat{\mathbf{y}})\;\mathrm{m}$$
and so, the average velocity during the 7 seconds is
$$\mathbf{v}_{avg} = (\frac{4}{7}\hat{\mathbf{x}} + \frac{3}{7} \hat{\mathbf{y}})\;\mathrm{\frac{m}{s}}$$
Clearly, if another particle had this constant velocity and started at the same initial point at the same time as the first particle, the two would reach the same final point at the same time.
On the other hand, the quantity
$$\frac{\mathbf{v}_f + \mathbf{v}_i}{2}$$
is an average of two velocities, which is not particularly useful or meaningful, not an average velocity which has a clear and useful meaning.
There are two special cases:
(1) In the case that the particle spends half of the elapsed time at a constant velocity $\mathbf{v}_1$ and spends the other half of the elapsed time at a constant velocity $\mathbf{v}_2$, then the average velocity is just the average of the two velocities.
(2) In the case that the particle has constant acceleration, the velocity increases linearly with time and so the displacement per unit time (and working in 1-D)
$$\Delta r = v_i + \frac{(v_f - v_i)}{2} = \frac{v_i + v_f}{2}$$
and thus, the average velocity is just the average of the initial and final velocities.