Electronic – the difference between ‘x’ and ‘plus’ symbol in the block diagram

You have missed the feedback path.
(R(s) - 2*C(S))*5(s+4)/ ((s+5)(s+8)) = C(s), and so on

Also, the closed loop transfer function vs open in case the feedback is 1

H(s) = F(s) / (1+F(s)); in your case the feedback is 2 so, you have to move it in the loop.

Lets start by labelling your first effort

We will call the bottom block \$ G_3 = \frac{G_1}{1+ G_1 \cdot H_1} \$

The top block \$ G_4 = \frac{G_2}{1+ G_2 \cdot H_2} \$

For convenience I've dropped the (s) notation from \$ y \$ and \$ u \$ since we know they are functions of s.

I've also added labels:
\$ w \$ is the input to \$ G_4 \$,
\$ x \$ is the output of \$ G_3 \$

Now lets simplify further

\$ y = x + w \cdot G_4 \$

\$ w = u - x \cdot H_1 \$

\$ x = u \cdot G_3 \$

We want \$ \frac{y}{u} \$

\$ y = u \cdot G_3 + w \cdot G_4 \$

\$ w = u - u \cdot G_3 \cdot H_1 \$

\$ y = u \cdot G_3 + (u - u \cdot G_3 \cdot H_1) \cdot G_4 \$

\$ y = u \cdot \left( G_3 + G_4 - G_3 \cdot G_4 \cdot H_1 \right) \$

If we now replace \$ G_3 \$ and \$ G_4 \$ with their definitions and these by their transfer functions (in \$ s \$) we have the transfer function for the entire system. Can you take it from here?