The equation in the orange image is for a single sample. If a sample is drawn from a normal distribution with mean $\mu$ and standard deviation $\sigma$, then $\bar{x}$ is normal with mean $\mu$ and variance $\sigma/\sqrt{n}$.

Hence $\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}$ will have a standard normal distribution. However, if we estimate $\sigma/\sqrt{n}$ from the sample, by the estimated standard error of the mean, $s/\sqrt{n}$, (where $s$ is the standard deviation of the sample), then

$\frac{\bar{x}-\mu}{std.err(\bar{x})}=\frac{\bar{x}-\mu}{s/\sqrt{n}}$ has a $t$ distribution with $n-1$ degrees of freedom.

The orange image is consistent with what I just said.

But then you said:

t-score given H_null = estimated_mean / (std.error/n)

This is wrong; in fact you have several things wrong there. Note that the standard error of the mean is $s/\sqrt{n}$, where $s$ is the standard deviation of the data.

The t-statistic in regression is slightly different (though analogous in form; its $t=\frac{b-\beta}{s.e.(b)}$ where $b$ is the estimated coefficient).

It's important to understand the problem here with the more basic case first.

The equation in the orange image is for a single sample. If a sample is drawn from a normal distribution with mean $\mu$ and standard deviation $\sigma$, then $\bar{x}$ is normal with mean $\mu$ and variance $\sigma/\sqrt{n}$.

Hence $\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}$ will have a standard normal distribution. However, if we estimate $\sigma/\sqrt{n}$ from the sample, by the estimated standard error of the mean, $s/\sqrt{n}$, (where $s$ is the standard deviation of the sample), then

$\frac{\bar{x}-\mu}{std.err(\bar{x})}=\frac{\bar{x}-\mu}{s/\sqrt{n}}$ has a $t$ distribution with $n-1$ degrees of freedom.

The orange image is consistent with what I just said.

But then you said:

t-score given H_null = estimated_mean / (std.error/sqrt(n))

This is not correct. Note that the standard error of the mean is $s/\sqrt{n}$, where $s$ is the standard deviation of the data.

The t-statistic in regression is slightly different (though analogous in form; its $t=\frac{b-\beta}{s.e.(b)}$ where $b$ is the estimated coefficient).

It's important to understand the problem here with the more basic case first.

The equation in the orange image is for a single sample. If a sample is drawn from a normal distribution with mean $\mu$ and standard deviation $\sigma$, then $\bar{x}$ is normal with mean $\mu$ and variance $\sigma/\sqrt{n}$.

Hence $\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}$ will have a standard normal distribution. However, if we estimate $\sigma/\sqrt{n}$ from the sample, by the estimated standard error of the mean, $s/\sqrt{n}$, (where $s$ is the standard deviation of the sample), then

$\frac{\bar{x}-\mu}{std.err(\bar{x})}=\frac{\bar{x}-\mu}{s/\sqrt{n}}$ has a $t$ distribution with $n-1$ degrees of freedom.

The orange image is consistent with what I just said.

But then you said:

t-score given H_null = estimated_mean / (std.error/n)

This is wrong; in fact you have several things wrong there. Note that the standard error of the mean is $s/\sqrt{n}$, where $s$ is the standard deviation of the data.

The t-statistic in regression is slightly different (though analogous in form; its $t=\frac{b-\beta}{s.e.(b)}$). It's important to understand the problem here first.

The equation in the orange image is for a single sample. If a sample is drawn from a normal distribution with mean $\mu$ and standard deviation $\sigma$, then $\bar{x}$ is normal with mean $\mu$ and variance $\sigma/\sqrt{n}$.

Hence $\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}$ will have a standard normal distribution. However, if we estimate $\sigma/\sqrt{n}$ from the sample, by the estimated standard error of the mean, $s/\sqrt{n}$, (where $s$ is the standard deviation of the sample), then

$\frac{\bar{x}-\mu}{std.err(\bar{x})}=\frac{\bar{x}-\mu}{s/\sqrt{n}}$ has a $t$ distribution with $n-1$ degrees of freedom.

The orange image is consistent with what I just said.

But then you said:

t-score given H_null = estimated_mean / (std.error/n)

This is wrong; in fact you have several things wrong there. Note that the standard error of the mean is $s/\sqrt{n}$, where $s$ is the standard deviation of the data.

The t-statistic in regression is slightly different (though analogous in form; its $t=\frac{b-\beta}{s.e.(b)}$ where $b$ is the estimated coefficient). 

It's important to understand the problem here with the more basic case first.