Let X be normally distributed random variable with expected value $\mu$ and standard deviation $\sigma$, then its СDF is: $$ F(x)=\frac{1}{\sigma\sqrt{2\pi}} \int_{-\infty}^x e^{\frac{(t-\mu)^{2}}{2\sigma ^{2}}} dt $$
So that normalize it to standard normal distribution we do this: $$ Z = \frac{X - \mu}{\sigma}, $$ where Z is standard normal random variable derived from X. This means that the CDF of Z is: $$ F(z)=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^z e^\frac{{t^{2}}}{2} dt $$ As I understand the foregoing the CDF of X means that $$ F(x) = P (X < x) $$ and CDF of Z means that $$ F(z) = P(\frac{X - \mu}{\sigma} < z) $$ therefore $$ F(z)=P(X < z\sigma + \mu) $$ And if we take this CDF: $$ F(x)=\frac{1}{\sigma\sqrt{2\pi}} \int_{-\infty}^{x\sigma + \mu} e^{\frac{(t-\mu)^{2}}{2\sigma ^{2}}} dt $$ then we get the standartized normal distribution of X.
Am I not correct?
Aside from the typos, yes that seems to be right. You just need to keep going to make it "look right".
Let $X\sim N(\mu, \sigma^2)$, and $Z = \frac{X-\mu}{\sigma}$. Then
\begin{align*} P(Z<z)& = P\left(\frac{X-\mu}{\sigma}<z\right)\\ &=P(X<\sigma z+\mu)\\ &=\int_{-\infty}^{\sigma z+\mu} \frac{1}{\sqrt{2\pi}\sigma}\exp\left\{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2\right\}\,dx\tag 1\\ &=\int_{-\infty}^{z}\frac{1}{\sqrt{2\pi}\sigma}\exp\left\{-\frac{1}{2}u^2\right\}\cdot\sigma\,du\\ &=\int_{-\infty}^{z}\frac{1}{\sqrt{2\pi}}\exp\left\{-\frac{1}{2}u^2\right\}\,du\\ &=\Phi(z) \end{align*} where $\Phi$ is the standard normal cdf, and in $(1)$ I made the subsitution $$u = \frac{x-\mu}{\sigma}.$$