It is my personal opinion that the vector addition the OP is unsatisfied with can not be significantly improved.
So what follows here is not really an answer but rather some comments.
What's confusing in the first place is the phrase
barycentric coordinates
in the title.
As has become clear from the context, coordinates of the barycenter are meant instead, but the terminology and the description in the tag (barycentric-coordinates) is somehow confusing indeed, therefore I think
the OP is not to be blamed for this.
If attention is restricted to a triangle, and if it is assumed that all weights
$\alpha$ are equal, then three cases can be distinguished:
- Vertex Barycenter: $\vec{v}_Z = \left(\vec{v}_1 + \vec{v}_2 + \vec{v}_3\right)/3$
- Area Barycenter: $\vec{v}_A = \iint_\Delta \vec{v}(x,y)\,dx\,dy\, / \iint_\Delta dx\,dy$
- Edge Barycenter: $\vec{v}_E = \oint_\Delta \vec{v}(t)\left|\dot{\vec{v}}(t)\right| dt
/ \oint_\Delta \left|\dot{\vec{v}}(t)\right| dt$
Working out for the Area Barycenter - and the $(\xi,\eta)$ employed here are
really,
really
barycentric coordinates:
$$\vec{v}(\xi,\eta) = \vec{v}_1 + (\vec{v}_2-\vec{v}_1)\xi+(\vec{v}_3-\vec{v}_1)\eta \\ \Longleftrightarrow
\quad \begin{cases} x = x_1+(x_2-x_1)\xi+(x_3-x_1)\eta\\y = y_1+(y_2-y_1)\xi+(y_3-y_1)\eta\end{cases}
$$
Giving:
$$
\vec{v}_A = \frac{
\vec{v}_1\Delta\iint d\xi\,d\eta + (\vec{v}_2-\vec{v}_1)\Delta\iint\xi\,d\xi\,d\eta + (\vec{v}_3-\vec{v}_1)\Delta\iint\eta\,d\xi\,d\eta
}{\Delta\iint d\xi\,d\eta}
$$
with Jacobian
$$
\Delta = \operatorname{det}(\vec{v}_2-\vec{v}_1,\vec{v}_3-\vec{v}_1) =
\begin{vmatrix} x_2-x_1 & x_3-x_1 \\ y_2-y_1 & y_3-y_1\end{vmatrix}
$$
and
$$
\iint\xi\,d\xi\,d\eta = \int_0^1\left[\int_0^{1-\xi} d\eta\right]\xi\,d\xi = \int_0^1 (1-\xi)\xi\,d\xi = \frac{1}{6}\\
\iint\eta\,d\xi\,d\eta = \int_0^1\left[\int_0^{1-\xi} \eta d\eta\right]\,d\xi = \int_0^1 (1-\xi)^2/2\,d\xi = \frac{1}{6}\\
\iint d\xi\,d\eta = \int_0^1\left[\int_0^{1-\xi} d\eta\right] d\xi = \int_0^1 (1-\xi) d\xi = \frac{1}{2}
$$
Finally resulting in:
$$
\vec{v}_A = \frac{\vec{v}_1\Delta/2 + (\vec{v}_2-\vec{v}_1)\Delta/6 + (\vec{v}_3-\vec{v}_1)\Delta/6}{\Delta/2} =
\left(\vec{v}_1 + \vec{v}_2 + \vec{v}_3\right)/3 = \vec{v}_Z
$$
Which is, entirely by "coincidence", exactly the same expression as with the Vertex Barycenter
(assuming that coincidence is something that does exist in mathematics, which I don't believe).
Anyway, the equivalence between Vertex Barycenter and the Area Barycenter explains why it is possible
to find the
former with a construct that belongs to the
latter, namely, as it is formulated in the question:
by taking the midpoint of each side and drawing a line to the opposite vertex. Indeed, because
then we have subdivided the triangle in two balanced equal areas each time.
For the sake of completeness, we shall work out the last case too, for the Edge Barycenter
$\vec{v}(t) = \vec{v}_a(t)\cup\vec{v}_b(t)\cup\vec{v}_c(t)$ , where:
$$\begin{cases}
\vec{v}_a(t) = \vec{v}_2 + \left(\vec{v}_3-\vec{v}_2\right)t \\
\vec{v}_b(t) = \vec{v}_3 + \left(\vec{v}_1-\vec{v}_3\right)t \\
\vec{v}_c(t) = \vec{v}_1 + \left(\vec{v}_2-\vec{v}_1\right)t \end{cases} \quad \mbox{with} \quad 0 \le t \le 1
$$
Resulting in:
$$
\vec{v}_E = \frac{\left|\vec{v}_3-\vec{v}_2\right|\left(\vec{v}_3+\vec{v}_2\right)/2
+\left|\vec{v}_1-\vec{v}_3\right|\left(\vec{v}_1+\vec{v}_3\right)/2
+\left|\vec{v}_2-\vec{v}_1\right|\left(\vec{v}_2+\vec{v}_1\right)/2}{
\left|\vec{v}_3-\vec{v}_2\right|+\left|\vec{v}_1-\vec{v}_3\right|+\left|\vec{v}_2-\vec{v}_1\right|}
$$
It is noticed that this is only equ(iv)al(ent) with the Vertex Barycenter if the lengths of the three edges of the triangle
are all equal, such in sharp contrast with the Area Barycenter result.
EDIT. The following is from the Wikipedia article about Quadrilaterals:
The "vertex centroid" is the intersection of the two bimedians. As with any polygon, the $x$ and $y$ coordinates of the vertex centroid
are the arithmetic means of the $x$ and $y$ coordinates of the vertices.
The "area centroid" of quadrilateral ABCD can be constructed in the following way. Let Ga, Gb, Gc, Gd be the centroids of triangles
BCD, ACD, ABD, ABC respectively. Then the "area centroid" is the intersection of the lines GaGc and GbGd.
It is thus confirmed that the Vertex Barycenter and the Area Barycenter, quite in general, are not the same for quadrilaterals. And likewise their geometrical construction.
Last but not least. The area centroid is not at all linear in the vertex coordinates for an arbitrary (convex) polygon, not even for an arbitrary (convex) quadrilateral:
Thus, in general :
$\vec{v}_A \ne \alpha_1 \vec{v}_1 + \alpha_2 \vec{v}_2 + \alpha_3 \vec{v}_3 + \alpha_4 \vec{v}_4$ .
