Both formulas are correct. Here is how you can get formula 2, by minimize the sum of squared errors.
$$
\sum^n_i(y^i-\hat{y^i})^2
$$
Dividing it by $n$ gives you MSE, and by $2$ gives you SSE used in the formula 2. Since you are going to minimize this expression with partial derivation technique, choosing $2$ makes the derivation look nice.
Now you use a linear model where $\hat{y} = f(h)$ and $h(x) = \theta{x}$, you get, (I omit the transpose symbol for $\theta$ in $\theta^T{x}$)
$$
\frac{1}{2}\sum^n_i(y^i-{f(\theta{x^i})})^2
$$
When you compute its partial derivative over $\theta$ for the additive term, you have,
$$
\frac{\partial}{\partial\theta_j}(\frac{1}{2}(y-{f(\theta{x})})^2)
= -(y-\hat{y}) f'(\theta{x}) x_j \quad (1)
$$
This is the formula 2 you gave. I don't give the detailed steps, but it is quite straightforward.
Yes, $f(h)$ is the activation function, and you do have the factor $f'(h)$ in the derivative expression as shown above. It disappears if it equals 1, i.e., $f(h) = h + c$, where $c$ is invariable w.r.t. $h$.
For example, if $h_\theta(x) = \theta{x}$, (i.e., $\sum^n_i{\theta_i{x_i}}$), and the prediction model is linear where $f(x) = \theta{x}$, too, then you have $f(h) = h$ and $f'(h) = 1$.
For another example, if $h_\theta(x) = \theta{x}$, while the prediction model is sigmoid where $f(h) = \frac{1}{1+e^-h}$, then $f'(h) = f(h)(1-f(h))$. This is why in book Artificial Intelligence: A Modern Approach, the derivative of logistic regression is:
$$
\frac{\partial}{\partial\theta_j}(\frac{1}{2}(y-{f(\theta{x})})^2) = -(y-\hat{y}) \hat{y}(1-\hat{y})x_j \quad (2)
$$
On the other hand, the formula 1, although looking like a similar form, is deduced via a different approach. It is based on the maximum likelihood (or equivalently minimum negative log-likelihood) by multiplying the output probability function over all the samples and then taking its negative logarithm, as given below,
$$
\sum^n_i-\log{P(y^i|x^i; \theta)}
$$
In a logistic regression problem, when the outputs are 0 and 1, then each additive term becomes,
$$
−logP(y^i|x^i;\theta) = -(y^i\log{h_\theta(x^i)} + (1-y^i)\log(1-h_\theta(x^i)))
$$
The formula 1 is the derivative of it (and its sum) when $h_\theta(x) = \frac{1}{1+e^{-\theta{x}}}$, as below,
$$
\frac{\partial}{\partial\theta_{j}}(−logP(y|x;\theta)) = -(y - \hat{y})x_j \quad (3)
$$
The derivation details are well given in other post.
You can compare it with equation (1) and (2). Yes, equation (3) does not have the $f'(h)$ factor as equation (1), since that is not part of its deduction process at all. Equation (2) has additional factors $\hat{y}(1-\hat{y})$ compared to equation (3). Since $\hat{y}$ as probability is within the range of (0, 1), you have $\hat{y}(1-\hat{y}) < 1$, that means equation (2) brings you a gradient of smaller absolute value, hence a slower convergence speed in gradient descent than equation (3).
Note the sum squared errors (SSE) essentially a special case of maximum likelihood when we consider the prediction of $\hat{y}$ is actually the mean of a conditional normal distribution. That is,
$$
p(y | x) = N(y;\hat{y},I).
$$
If you want to get more in depth knowledge in this area, I would suggest the Deep Learning book by Ian Goodfellow, et al.
