I don't know the answer, but I can at least provide some context that may convince others that this question is actually something that is studied and is indeed hard.

First, let's rewrite the question so it's easier to digest. We'll first define $Z := V{\rm diag}(x_1,x_2,0,0)V^*$ for convenience. Notice that the condition $x_1|v_{k,1}|^2 + x_2|v_{k,2}|^2 = 0$ for $1 \leq k \leq 4$ is equivalent to saying that $Z$ has zeroes on its diagonal.

Also, $U_i{\rm diag}(1,0,0,0)U_i^* = u_i u_i^*$, where $u_i$ is the first column of $U_i$, so the condition ${\rm tr}(U_i{\rm diag}(1,0,0,0)U_i^*V{\rm diag}(x_1,x_2,0,0)V^*) = 0$ for $1 \leq i \leq 7$ is equivalent to ${\rm tr}(Zu_iu_i^*) = 0$ for $1 \leq i \leq 7$.

So a rephrasal of the question is: given unit vectors $\{u_i\}_{i=1}^7$, does there exist a $4\times 4$ Hermitian matrix $Z$ with rank $2$ such that ${\rm tr}(Zu_iu_i^*) = 0$ for $1 \leq i \leq 7$ and ${\rm tr}(Ze_ie_i^*) = 0$ for $1 \leq i \leq 4$, where $\{e_i\}_{i=1}^4$ is the standard basis of $\mathbb{R}^4$?

In other words, we are given eleven rank-$1$ positive semidefinite $4 \times 4$ matrices ($\{u_iu_i^*\}_{i=1}^7$ and $\{e_ie_i^*\}_{i=1}^4$), and we are asking whether or not there is always a rank-$2$ Hermitian matrix orthogonal to all of them. This is the central question studied in arXiv:1109.5478 -- it is a question about how many quantum measurements are necessary to reconstruct a pure quantum state (i.e., a rank-$1$ positive semidefinite matrix).

Anyway, if you trace through their results, you can see that the answer to your question is "no" if you have eight unitary matrices $U_i$ rather than seven. I'm not sure about the 7-unitary case.

I don't know the answer, but I can at least provide some context that may convince others that this question is actually something that is studied and is indeed hard.

First, let's rewrite the question so it's easier to digest. We'll first define $Z := V{\rm diag}(x_1,x_2,0,0)V^*$ for convenience. Notice that the condition $x_1|v_{k,1}|^2 + x_2|v_{k,2}|^2 = 0$ for $1 \leq k \leq 4$ is equivalent to saying that $Z$ has zeroes on its diagonal.

Also, $U_i{\rm diag}(1,0,0,0)U_i^* = u_i u_i^*$, where $u_i$ is the first column of $U_i$, so the condition ${\rm tr}(U_i{\rm diag}(1,0,0,0)U_i^*V{\rm diag}(x_1,x_2,0,0)V^*) = 0$ for $1 \leq i \leq 7$ is equivalent to ${\rm tr}(Zu_iu_i^*) = 0$ for $1 \leq i \leq 7$.

So a rephrasal of the question is: given unit vectors $\{u_i\}_{i=1}^7$, does there exist a $4\times 4$ Hermitian matrix $Z$ with rank $2$ such that ${\rm tr}(Zu_iu_i^*) = 0$ for $1 \leq i \leq 7$ and ${\rm tr}(Ze_ie_i^*) = 0$ for $1 \leq i \leq 4$, where $\{e_i\}_{i=1}^4$ is the standard basis of $\mathbb{R}^4$?

In other words, we are given eleven Hermitian $4 \times 4$ matrices ($\{u_iu_i^*\}_{i=1}^7$ and $\{e_ie_i^*\}_{i=1}^4$), and we are asking whether or not there is always a rank-$2$ Hermitian matrix orthogonal to all of them. This is the central question studied in arXiv:1109.5478 -- it is a question about how many quantum measurements are necessary to reconstruct a pure quantum state.

Anyway, they show that the smallest set of $4 \times 4$ Hermitian matrices with the property that every orthogonal matrix has rank $\geq 3$ has $9$ elements. They also show that almost every set of $13$ or more $4 \times 4$ Hermitian matrices have the property that every orthogonal matrix has rank $\geq 3$. You have given a set that lies somewhere in the middle (it consists of $11$ matrices, but those matrices are promised to be rank $1$ and four of them are mutually orthogonal), so it's not immediately clear to me what the answer is.