Let $K \subset \mathbb{R}^d$ ($d \in \mathbb{N}$) be a nonempty compact set, and
$K_{\varepsilon}= \{ x \in \mathbb{R}^d;d(x, K)< \varepsilon \}$, where $\displaystyle d(x, K)= \inf_{y \in K} |x-y|$.
Assume that $\forall x \in \mathbb{R}^d, \exists1 f(x) \in K$ such that
$d(x, K)= |x -f(x)|$.
Then does the following hold?
$\int_{K_1} g(f(x)+t(x-f(x))) dx= \frac{{\rm Vol}(K_1)}{{\rm Vol}(K_t)}\int_{K_{t}} g(y)dy \ $ for all $t>0, g \in C^{\infty}_0 (\mathbb{R}^d)$?
No, this isn't true. For instance, suppose $K$ is a closed ball and $g$ is a bump function whose support is contained in the interior of $K$. Then both your integrals will be just the integral of $g$ over its support, and so the integrals are equal, even though $K_1$ and $K_t$ have different volume.
More generally, your equation would be correct if the change of variables $y=f(x)+t(x-f(x))$ scaled volumes inside $K_1$ by the constant factor $\frac{{\rm Vol}(K_1)}{{\rm Vol}(K_t)}$. But it doesn't: inside $K$ it fixes volumes, and between $K$ and $K_1$ it scales them in a complicated nonconstant way (at least if $d>1$, so that the surface areas of the sets $\{x:d(x,K)=r\}$ are not constant in $r$).