Time evolution in quantum mechanics is usually done with the so-called time evolution operator,
$$
U(t_f,t_i)=\text{e}^{-\text{i}H(t_f-t_i)/\hbar},
$$
such that, when applied on a wave function at initial time $t=t_i$,
$$
U(t_f,t_i)\, \psi(x,t_i) = \psi(x, t_f),
$$
the wave function at the final time $t=t_f$ follows. For a free particle, the Hamiltonian reads,
$$
H=\frac{p^2}{2m},
$$
so we get the wave function for $t=t_f$ if we let $H$ act on $\psi(x)$. Before we can do that, we have to express $p$ as a derivative acting on $x$,
$$
p=\hbar k \to -\text{i}\hbar \partial_x\quad\Rightarrow\quad H=-\frac{\hbar^2}{2m}\partial_x^2
$$
(note $\partial_x \equiv \tfrac{\partial}{\partial x}$) so that we write,
$$
\psi(x, t_f) = \text{e}^{\text{i}\frac{\hbar}{2m}\partial_x^2(t_f-t_i)}\, \psi(x,t_i).
$$
According to your question, we will use $t_i=0$ and $t_f=t$:
$$
\psi(x, t) = \text{e}^{\text{i}\frac{\hbar}{2m}\partial_x^2t}\, \psi(x, 0).
$$
I'm not exactly sure why, but in order to match your image, we now do a Fourier transform on the initial wave function,
$$
\psi(x,0) = \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty \text{d}k\, \text{e}^{ikx}\, \phi(k).
$$
With this, we arrive at the equation
$$
\begin{align*}
\psi(x, t) &= \text{e}^{\text{i}\frac{\hbar}{2m}\partial_x^2 t} \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty \text{d}k\, \text{e}^{ikx}\, \phi(k)\\
&= \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty \text{d}k\, \text{e}^{\text{i}\frac{\hbar}{2m}\partial_x^2 t} \text{e}^{ikx}\, \phi(k)\\
&= \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty \text{d}k\, \text{e}^{\text{i}\frac{\hbar}{2m}(\text ik)^2 t} \text{e}^{ikx}\, \phi(k)\\
&= \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty \text{d}k\, \text{e}^{-\text{i}\frac{\hbar k^2}{2m} t} \text{e}^{ikx}\, \phi(k)\\
&= \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty \text{d}k\, \text{e}^{i\left( kx -\frac{\hbar k^2}{2m}t\right)}\, \phi(k).
\end{align*}
$$
The third equality follows by an implicit Taylor expansion and letting every power of $\partial_x$ act on $\text e^{\text ikx}$, see e.g. here or here for details.
So, to answer your question: $\phi(k)$ is only a function of $k$, because we did the Fourier transform only on $\psi(x)$. If we were to do a Fourier transform on $\psi(x,t)$, we would get a $\phi(k,\omega)$.