There are probably tricks one can use to solve this particular problem, but one can solve this sort of problem in general by using Gröbner bases. I highly recommend Cox, Little, and O'Shea's Ideals, Varieties, and Algorithms: they treat this exact problem in Chapter 3 Elimination, particularly in $\S3$ Implicitization. The main result is the following theorem.
Theorem 1 (Polynomial Implicitization)
If $k$ is an infinite field, let $F: k^m \to k^n$ be the function determined by the polynomial parametrization
\begin{align*}
x_1 &= p_1(t_1, \ldots, t_m)\\
&\ \, \vdots\\
x_n &= p_n(t_1, \ldots, t_m) \, .
\end{align*}
Let $I = \langle x_1 - p_1, \ldots, x_n - p_n \rangle$ and let $I_m = I \cap k[x_1, \ldots, x_n]$ be the $m^\text{th}$ elimination ideal. Then $\mathbb{V}(I_m)$ is the smallest variety in $k^n$ containing $F(k^m)$.
This deals with affine varieties, and the results in $\S$5 of Chapter 8 can be used to extend to projective varieties.
Let $z_0, \ldots, z_5$ be the coordinates on $\mathbb{P}^5$. The image of $v$ is given parametrically by the equations
$$
z_0 = x_0^2 \qquad z_1 = x_1^2 \qquad z_2 = x_2^2 \qquad z_3 = x_0 x_1 \qquad z_4 = x_0 x_2 \qquad z_5 = x_1 x_2 \, .
$$
Let $I = \langle z_0-x_0^2, z_1-x_1^2, z_2-x_2^2, z_3-x_0 x_1, z_4-x_0 x_2, z_5-x_1 x_2 \rangle \trianglelefteq k[x_0,x_1,x_2,z_0, \ldots, z_5]$ be the ideal corresponding to this parametrization. I computed a Gröbner basis for $I$ in Sage with respect to the graded lexicographic monomial ordering $x_0 > x_1 > x_2 > z_0 > \cdots > z_5$ as follows:
R.<x0,x1,x2,z0,z1,z2,z3,z4,z5> = PolynomialRing(QQ,9,order = "deglex")
gens = [z0-x0^2, z1-x1^2, z2-x2^2, z3-x0*x1, z4-x0*x2, z5-x1*x2]
I = Ideal(gens)
G = I.groebner_basis()
This outputs a Gröbner basis containing $20$ generators. Those only involving the $z_i$ are
$$
z_0 z_1 - z_3^2, \qquad z_0z_2 - z_4^2, \qquad z_0 z_5 - z_3 z_4, \qquad z_1 z_2 - z_5^2, \qquad z_1 z_4 - z_3 z_5
$$
and these are the quadrics defining the image of $v$.
I will use the following notation The definition of the Veronese map will be taken from the question above. I will denote the coordinates in $\mathbb{P}^{5}$ by $[z_{0}:\ldots:z_{5}]$ and the coordinates in $\mathbb{P}^{2}$ by $[x_{0}:x_{1}: x_{2}]$.
Claim The Veronese surface is cut out by three quadrics:
$$C_{1}: z_{0}z_{3} - z_{1}^{2} = 0$$
$$C_{2}: z_{0}z_{5} -z_{2}^{2}= 0 $$
$$C_{3}: z_{3}z_{5} - z_{4}^{2} = 0. $$
It should be checked that they are indeed enough to define the locus but we will not get into that.
We consider a quadric $Q$ in $\mathbb{P}^{2}$ given by $Q: a_{0}x^{2}_{0} -(a_{1}x^{2}_{1} + a_{2}x^{2}_{2}) = 0 $. We can and will assume that $a_{0} \neq 0$ and after rescaling $a_{0} = 1$.
Clearly the image of $Q$ is contained in the hyperplane $H: z_{0} - a_{1}z_{3} - a_{2}z_{5} = 0$.
Our goal: Understand the intersection of this hyperplane with quadrics $C_{i}$.
$$C_{1} \cap H : (a_{1}z_{3} + a_{2}z_{5})z_{3}-z_{1}^{2}= 0$$
$$C_{2} \cap H : (a_{1}z_{3} + a_{2}z_{5})z_{5}-z_{2}^{2}= 0$$
Multiplying the first equation above by $z_{5}$ and the second one by $z_{3}$ and subtracting them we get the equation $E:z_{1}^{2}z_{5} - z_{2}^{2}z_{3} = 0.$
This equation $E$ represents the locus of the intersection of $H$ with two of the three quadrics. The justification for this ad-hoc process is the the following:
Inside the open (quasi-affine) variety $z_{3}z_{5} \neq 0$ we are allowed to multiply by non-zero functions $z_{3}$ and $z_{5}$. Outside this locus one has to check that this equation is still valid.
Next we study the intersection of $E$ and $C_{3}$. Again one multiplies $E$ by $z_{3}$ and $C_{3}$ by $z_{1}^{2}$ subtract the resulting equations and get $z_{1}^{2}z_{4}^{2}- z_{2}^{2}z_{3}^{2} = 0$ - a quartic.
The computation of the image of a cubic in $\mathbb{P}^{2}$ under the Veronese in Harris's book, page 2 here , is a great illustration of the complexity of figuring out the intersections.
Best Answer
A general hyperplane section of the Igusa quartic threefold $$ \Bigg\{ \sum_{i=1}^6 x_i = 4\Big(\sum_{i=1}^6 x_i^4\Big) - \Big(\sum_{i=1}^6 x_i^2\Big)^2 = 0 \Bigg\} \subset \mathbb{P}^5 $$ has 15 ordinary double points.