I don't know your purpose, but here's some not-probably-sharp estimate that works for any $k$. 
Put $\rho=\frac{1}{d}\|A|_{({\mathbb C}1)^\perp}\|$ and $\gamma$ to be the positive root of $t^2-\rho t -\rho=0$. 
One has $\gamma<\sqrt{2\rho}<1$ when $\rho<\frac{1}{2}$. 
Then for any $f$ with $\sum f(v)=0$ and $\|f\|_\infty\le1$, one has 
$$\frac{1}{|V|\cdot d^{k-1}}\left|\sum_{v_1,v_2,\ldots,v_k : \{v_i,v_{i+1}\}\in E}
  f(v_1)\cdots f(v_k)\right| \le \gamma^k.$$

Proof. For $D:=\mathrm{diag}\,f \in B(\ell_2V)$ and $B:=\frac{1}{d}AD$, 
the LHS is $\frac{1}{|V|}|\langle B^{k-1}1_V,f\rangle|$.
With respect to the orthogonal decomposition $\ell_2V={\mathbb C}1_V\oplus ({\mathbb C}1_V)^\perp$, 
one writes $B$ as an operator matrix
$B=\left[\begin{smallmatrix} 0 & b \\ c & d \end{smallmatrix}\right]$, 
where $\| b\|\le 1$, $\|c\|\le\rho$, and $\|d\|\le\rho$. 
Hence for 
$C:=\left[\begin{smallmatrix} 0 & 1 \\ \rho & \rho \end{smallmatrix}\right]
 \in M_2({\mathbb R})$ with the eigenvalue $\gamma>0$ and 
the eigenvector 
$\left[\begin{smallmatrix} 1 \\ \gamma \end{smallmatrix}\right]$, 
one gets
$$\frac{1}{|V|}|\langle B^{k-1}1_V,f\rangle|
 \le \left[\begin{smallmatrix} 0 & 1 \end{smallmatrix}\right] C^{k-1} 
  \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]
 \le \left[\begin{smallmatrix} 0 & 1 \end{smallmatrix}\right] C^{k-1} 
  \left[\begin{smallmatrix} 1 \\ \gamma \end{smallmatrix}\right]=\gamma^k.$$

It's probably worth noting that the same proof shows 
$$\frac{1}{|V|}\sum_{v_1\in V}\left|\frac{1}{d^{k-1}}\sum_{v_2,\ldots,v_k : \{v_i,v_{i+1}\}\in E}
  f_1(v_1)\cdots f_k(v_k)\right|^2 \le 2\gamma^{2(k-1)}.$$