Let $\mu_{x,y;t}$ be the Wiener measure generated by $\exp[t \Delta](x,y)$. I see in my book the following step: for any $0\leq s \leq t$,
$$\int dx\int d\mu_{x,x;t}(\omega) \phi(x) = \int dx\int d\mu_{x,x;t}(\omega) \phi(\omega(0)) = \int dx \int d\mu_{x,x;t}(\omega) \phi(\omega(s)) $$
The first '=' comes from $\omega(0) = x $. The second '=' comes from the fact that $\exp[t \Delta](x,y)$ forms a semi group.
However, I don't understand the second '='. How can I see this?
The question is unclear, here is an interpretation. In the $\mathbb R^d$ case, let us assume that $\mu_{x,y;t}$ is the Brownian bridge probability measure on $\Omega=C([0,t],\mathbb R^d)$. In other words, for each $x$ in $\mathbb R^d$ and $t\gt0$, $W_{x,t}$ is the Wiener measure on $\Omega$ of the Brownian motion $(B(s))_{0\leqslant s\leqslant t}$ of length $t$ starting from $B(0)=x$, and the family $(\mu_{x,y;t})_{y\in\mathbb R^d}$ is the decomposition of $W_{x,t}$ according to the value of $B(t)$, that is, $\mu_{x,y;t}(B(t)=y)=1$ for every $y$, and $$ W_{x,t}(\,\cdot\,)=\int_{\mathbb R^d}\mu_{x,y;t}(\,\cdot\,)\,\mathrm d\nu_{x,t}(y), $$ where $\nu_{x,t}$ is the distribution of $B(t)$ with respect to $W_{x,t}$. Consider the integral on the left of your double identity, namely, $$ A_\varphi=\int_{\mathbb R^d}\mathrm dx\int_\Omega\mathrm d\mu_{x,x:t}(\omega)\varphi(x), $$ then $$ A_\varphi=\int_{\mathbb R^d}\mathrm dx\varphi(x)\int_\Omega\mathrm d\mu_{x,x:t}(\omega)=\int_{\mathbb R^d}\mathrm dx\varphi(x), $$ because $\mu_{x,x:t}$ is a probability measure. Likewise, the integral in the middle is $$ \int_{\mathbb R^d}\mathrm dx\int_\Omega\mathrm d\mu_{x,x:t}(\omega)\varphi(\omega(0))=\int_{\mathbb R^d}\mathrm dx\int_\Omega\mathrm d\mu_{x,x:t}(\omega)\varphi(x)=A_\varphi, $$ because $\omega(0)=x$ almost surely for $\mu_{x,x;t}$. Finally, for every $s$ in $[0,t]$, the integral on the right is $$ C_\varphi=\int_{\mathbb R^d}\mathrm dx\int_\Omega\mathrm d\mu_{x,x:t}(\omega)\varphi(\omega(s)). $$ Note that $\omega(s)$ under $\mu_{x,x:t}$ is distributed like $x+\omega(s)$ with respect to $\mu_{0,0;t}$ hence, using Fubini theorem, $$ C_\varphi=\int_\Omega\mathrm d\mu_{0,0:t}(\omega)\int_{\mathbb R^d}\mathrm dx\varphi(x+\omega(s))=\int_\Omega\mathrm d\mu_{0,0:t}(\omega)\int_{\mathbb R^d}\mathrm dz\varphi(z), $$ by the change of variable $x=z-\omega(s)$, and finally, $$ C_\varphi=A_\varphi. $$ In the end, the crucial argument is the way the measures $\mu_{x,y;t}$ behave under the action of translations. No idea why the semi-group property should be involved here, or how the argument above would carry to the case of manifolds though.