A dominate vector subset sum problem
It's hard for $k\ge 2$ by a reduction from Partition. Let's first look at $k = 3$. Suppose that the input to Partition is the numbers $x_1, \ldots, x_n$, and their sum is $S$. For each $i$ create a vector $v_i$ whose first coordinate is $x_i$, second coordinate is $-x_i$, and the third coordinate is $0$. Add another vector $v_{n+1}$ with first coordinate $0$, second coordinate $S$, and last coordinate $1$. Define $t$ to have first and second coordinates $S/2$, and last coordinate $1$. Then any solution $I$ to this instance of your problem must include $n+1$, and the sum $\sum_{i \in I}{v_i}$ has first coordinate $\sum_{i \in I\setminus \{n+1\}}{x_i}$ and second coordinate $\sum_{i \in [n] \setminus I}{x_i}$, so it's only feasible if $I \setminus \{n+1\}$ is a valid solution to the Partition instance.

This can be extended to $k=2$ with a trick. Define the same instance but $v_i$ for $i \in [n]$ has only the first two coordinates. Let $A = \sum_{i=1}^n{|x_i|}$ and $b = \lceil \log_2 A\rceil$. Define $v_{n+1}$ to have first coordinate $2^b$, and second coordinate $S$, and $t$ to have first coordinate $2^b+S/2$ and second coordinate $S/2$. This forces the $b$-th bit (from the right) of the first coordinate of $\sum_{i \in I}{v_i}$ to be $1$, which can only happen if $n+1 \in I$. The rest is the same as above.