Complex-doubly periodic function in two variables?

The answer is that the only solutions have the form
$$f = (f_1,f_2) = \bigl(c, h(\,\overline{z}_1, z_2)\bigr)$$
where $h:\mathbb{C}^2\to\mathbb{C}$ is holomorphic and $c$ is a constant,
which must equal zero unless $k_1$ and $k_2$ are integers.

The argument is as follows: The first equation implies that there exists a function $g:\mathbb{C}^2\to\mathbb{C}$ such that
$$f_1 = \frac{\partial g}{\partial z_1} \quad\text{and}\quad f_2 = -\frac{\partial g}{\partial z_2}.$$
Substituting this into the second equation implies that $g$ must satisfy
$$\frac{\partial^2 g}{\partial z_1\partial\overline{z}_1} + \frac{\partial^2 g}{\partial z_2\partial\overline{z}_2} = 0.$$
In other words $g$ is a harmonic function on $\mathbb{C}^2$. Since $g$ is harmonic, so is its derivative with respect to $z_1$, i.e., $f_1$.

The periodicity conditions imposed on $f_1$ imply that $f_1$ is bounded, and a bounded harmonic function on $\mathbb{C}^2$ is constant. Thus, $f_1 = c$ for some constant $c\in\mathbb{C}$. Obviously, $c$ must be zero unless $k_1$ and $k_2$ are integers.

Since $f_1$ is constant, the given equations on $f_2$ reduce to
$$\frac{\partial f_2}{\partial z_1} = \frac{\partial f_2}{\partial\overline{z}_2} = 0.$$
Hence $f_2 = h(\overline{z}_1,z_2)$ for some homorphic function $h:\mathbb{C}^2\to\mathbb{C}$.

Remark: It wasn’t clear from the OP’s question whether the OP wanted $f$ to be ‘doubly-periodic’ or just $f_1$, nor was it clear exactly what the OP meant by ‘doubly-periodic’ because, normally, the ‘doubly-periodic’ condition wouldn’t have the exponential factors in its definition.