Let $(X_t)$ a solution of $$dX_t=b(X_t)dt+\sigma (X_t)dB_t,\quad X_s=x,\quad t\geq s.$$
Why $$\int_s^{s+h}\sigma (X_u^{s,x})dB_u=\int_0^h \sigma (X_{u+s}^{s,x})d\tilde B_u$$ where $\tilde B_u=B_{u+s}-B_u$ ?
For me, doing a substitution we have$$\int_s^{s+h}\sigma (X_u^{s,x})dB_u=\int_0^h \sigma (X_{u+s}^{s,x})dB_{u+s},$$ so I don't understand this $\tilde B_u=B_{u+s}-B_u$.
The identity $$\int_0^h \sigma(X_{u+s}^{s,x}) \, dB_{u+s} = \int_0^h\sigma(X_{u+s}^{s,x}) \, d\underbrace{(B_{s+u}-B_s)}_{=\tilde{B}_u} \tag{1}$$
holds because the stochastic integral is defined in terms of increments of the driving process.
Let's consider first the classical Riemann-Stieltjes integral: if you have some (nice) functions $f,g$ then $$\int_0^T f(u) \, dg(u) = \int_0^T f(u) \, d\tilde{g}(u)$$ for $\tilde{g}(u) := g(u)-c$ and $c \in \mathbb{R}$ is some constant. Analogously, we get $$\int_0^T f(s+u) \, dg(s+u) = \int_0^T f(s+u) \, d\tilde{g}(s+u)$$ for fixed $s \geq 0$. The identity $(1)$ is the corresponding statement for stochastic integrals - the only difference is that the constant $c=c(\omega)$ is now random.
To prove $(1)$ note that for "nice" functions $f$,
$$\int_0^t f(u+s) \, dB_{u+s} = \lim_{n \to \infty} \sum_{j=1}^{n} f(s+t_j) (B_{s+t_j}-B_{s+t_{j-1}})$$ where $t_j := tj/n$ and the limit is taken in $L^2(\mathbb{P})$. Since $$B_{s+t_j}-B_{s+t_{j-1}} = (B_{s+t_j}-B_s) -(B_{s+t_{j-1}}-B_s) = \tilde{B}_{t_j}-\tilde{B}_{t_{j-1}}$$
this gives
$$\int_0^t f(u+s) \, dB_{u+s} = \lim_{n \to \infty} \sum_{j=1}^{n} f(s+t_j) (\tilde{B}_{t_j}-\tilde{B}_{t_{j-1}}) = \int_0^t f(s+u) \, d\tilde{B}_u.$$