I have a rather natural idea for a construction of a differentiable atlas on $N$. But I don't know how to prove that it's unique. I call the atlas on $M$ "$A_M$".
Define the following atlas on $N$:
$A_N = \{ (u,\phi)\ |\ (f^{-1}(u),\phi \circ f) \in A_M \}$
To prove that this set is indeed an atlas, as well as differentiable, we start by showing that it 'covers' all of $N$:
Say $x\in N$. Then there will exist a chart $(v,\pi)$ in the atlas $A_M$ of $M$ containing $f^{-1}(x)\in M$. We claim that $(f(v), \psi\circ f^{-1})$ is an element of $A_N$. This is clearly the case, as $(f^{-1}(f(v)), \psi\circ f^{-1}\circ f)$ is an element of $A_M$. Hence the atlas contains a chart that has our arbitrarily chosen $x\in N$ in its domain. It therefore 'covers' all of $N$.
Let's show that the chart transition maps of $A_N$ are differentiable.
Assume $(u,\phi)$ and $(v,\psi)$ are charts in $A_N$ with $u\cup w\neq\emptyset$. Denote $u\cup w$ by $v$. Is the transition map
$t=\psi\circ\phi^{-1},\quad t:\phi(v)\rightarrow\psi(v)$
differentiable? Well, we know that $(\psi \circ f) \circ {(\phi \circ f)}^{-1}$ is differentiable, as $\psi \circ f$ and $\phi \circ f$ are chart maps in $A_M$. But we have:
$(\psi \circ f) \circ {(\phi \circ f)}^{-1} = (\psi \circ f) \circ (f^{-1} \circ \phi^{-1}) = \psi \circ (f \circ f^{-1}) \circ \phi^{-1} = \psi\circ\phi^{-1} = t$
Therefore $t$ is differentiable. So $A_N$ is a differentiable structure on $N$.
It is customary that the chart domains of an atlas are open sets. If we look at the definition of the set $A_N$ this is seen trivially for $A_N$: As $(f^{-1}(u),\phi \circ f)$ in $A_M$, $f^{-1}(u)$ is open in $M$. Since $f$ is homeomorphic, $u$ is therefore open in N. As this is the only instance were we rely on the fact that $f$ is homeomorphic, we would already be able to construct a differentiable structure on $N$ if we were only given the data of a continuous function $g:A_N\rightarrow A_M$.
Now, is $f:(M,A_M)\rightarrow (N,A_N)$ differentiable? Let $x\in M$ and $(u,\phi)$ be a chart in $A_M$ containing $x$. Then $(f(u),\phi\circ f^{-1})$ is a chart in $A_N$ containing $f(x)$. So, if we look at $f$ as a real function through these charts, is it differentiable? I.e., is $(\phi)\circ f\circ ({(\phi\circ f)}^{-1})$ differentiable? Well,
$(\phi)\circ f\circ ({(\phi\circ f)}^{-1}) = (\phi)\circ f\circ (f^{-1}\circ\phi^{-1}) = \phi\circ (f\circ f^{-1})\circ\phi^{-1} = \phi\circ\phi^{-1} = id_{\phi(u)}$.
Therefore $f$ is differentiable as a function between the differentiable manifolds $(M,A_M)$ and $(N,A_N)$.