No, the statement is false without anymore assumptions; you should just upgrade all your regularity assumptions right from the beginning. The idea of the counterexample is to work with endpoints of a graph of a bad-enough function and show that any smooth enough curve through those endpoints must exit the graph. Then, we invoke a standard lemma about the existence of smooth functions with a prescribed level-set.
The difference between this negative result and the result you’re alluding to in the comments (that for a smooth manifold, continuous path-connectedness is equivalent to smooth path-connectedness) is that here, we haven’t made any assumptions about $df$. I’m not sure at the moment, what a positive result would look like/ whether there even is a nice enough result with further assumptions on $df$ — but it is clear we need some extra information about the directions along which $f$ increases; this is needed to prohibit the above type of counterexample.
Lemma 1.
The graph $G$ of a continuous function $\phi:[0,1]\to\Bbb{R}$ is closed in $\Bbb{R}^2$.
The proof is trivial.
Lemma 2.
Let $\phi:[0,1]\to\Bbb{R}$ be a continuous, nowhere-differentiable function. Then, for any non-empty open interval $I\subset\Bbb{R}$ and $C^k$ curve $\gamma:I\to\Bbb{R}^2$ with $k\geq 1$, if we have $\text{image}(\gamma)\subset G$ (the graph of $\phi$) then $\gamma$ must be constant.
To prove this, suppose for contradiction that $\gamma$ is not constant. Then, there exist $t_1,t_2\in I$ such that $\gamma(t_1)\neq \gamma(t_2)$. Writing $\gamma = (\gamma_1,\gamma_2)$, we have that $(\gamma_1(t_1),\gamma_2(t_1))\neq (\gamma_1(t_2),\gamma_2(t_2))$; however, keeping in mind that the image of $\gamma$ lies inside a graph, it actually follows that $\gamma_1(t_1)\neq \gamma_1(t_2)$ (i.e if the first coordinates were equal, then because the points lie in a graph, the second coordinates must be equal as well). By the mean-value theorem, there must therefore exist a number $t_0$ in the open interval between $t_1$ and $t_2$ such that $\gamma_1’(t_0)\neq 0$. By the inverse function theorem (which can be applied since $\gamma_1$ is atleast $C^1$; in fact the IFT is pretty easy in 1D) it follows there is an open interval $U$ around $t_0$ in $I$ and an open interval $V\subset\Bbb{R}$ such that the restriction $\gamma_1:U\to V$ is a $C^k$ diffeomorphism.
The condition that the image of $\gamma$ lie in $G$ means that $\gamma_2=\phi\circ\gamma_1$. Restricting this to $U$, we see that $\gamma_2|_U= \phi|_V\circ (\gamma_1|_U)$. Inverting gives $\phi|_V=(\gamma_2|_U)\circ (\gamma_1|_U)^{-1}$, and the RHS is a composition of $C^k$ functions so is $C^k$. This implies $\phi$ is $C^k$ on a non-empty open interval $V$, contradicting that $\phi$ is nowhere-differentiable. Hence, it follows $\gamma$ must have been constant.
Lemma 3.
For any closed subset $C\subset\Bbb{R}^n$ there is a smooth function $f:\Bbb{R}^n\to\Bbb{R}$ such that $f$ vanishes on $C$ and $f$ is strictly positive on the complement of $C$.
This is a typical exercise with bump functions.
Finally, we come to the counterxample.
Theorem.
There is a smooth function $f:\Bbb{R}^2\to\Bbb{R}$, and distinct points $x,y\in\Bbb{R}^2$ such that for every continuous curve $\gamma:[0,1]\to\Bbb{R}^2$ which is $C^k$ on $(0,1)$ (for some $k\geq 1$) and such that $\gamma(0)=x$ and $\gamma(1)=y$, we have that $f\circ\gamma$ is NOT non-decreasing on $[0,1]$.
To prove this, fix a continuous, nowhere-differentiable function $\phi:[0,1]\to\Bbb{R}$, let $x= (0,\phi(0))$ and $y=(1,\phi(1))$, and let $G$ be the graph of $\phi$. By Lemmas 1 and 3, there is a smooth function function $f:\Bbb{R}^2\to\Bbb{R}$ such that $f|_G=0$ and $f|_{G^c}>0$. Now, let $\gamma:[0,1]\to \Bbb{R}^2$ be any continuous curve which is $C^k$ on $(0,1)$ and such that $\gamma(0)=x$ and $\gamma(1)=y$.
I claim there is a $t_0\in (0,1)$ such that $\gamma(t_0)\notin G$. If this were not true, then it means $\text{image}(\gamma|_{(0,1)})\subset G$ and hence by Lemma 2, $\gamma$ must be constant on $(0,1)$. By continuity, $\gamma$ is thus constant on $[0,1]$, which means $x=y$; but this is absurd since they’re different points on the graph. So, we thus have that
\begin{align}
(f\circ\gamma)(0)=f(x)=0,\quad\text{and}\quad (f\circ\gamma)(t_0)>0,\quad\text{and} \quad (f\circ\gamma)(1)=f(y)=0,
\end{align}
which shows that $f\circ\gamma$ starts at $0$, increases to a positive value, then decreases to $0$, and thus cannot be non-decreasing on $[0,1]$.
(But of course there is a continuous curve $\gamma:[0,1]\to\Bbb{R}^2$ through $x$ and $y$ along which $f$ is non-decreasing. Simply consider $\gamma(t)=(t,\phi(t))$, the natural parametrization of a graph; then $f\circ \gamma=0$ identically).
