You are missing an important piece of assumption here, which is that you have to consider an active transformation of the fields in order to achieve your goal of finding the stress-energy tensor. In your notation, that is to say
$$\widetilde{\phi}^{(i)}(\widetilde{x}) = \phi^{(i)} (x) \,. \tag7$$
That means $ \, \delta \phi^{(i)} = 0 $, implying that the conserved Noether current is
$$ J^\mu = -\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi^{(i)} )} \partial_\nu \phi^{(i)} \delta x^\nu + \mathcal{L} \delta x^\mu + F^\mu = {\Theta^\mu}_\nu \delta x^\nu + F^\mu \tag8$$
where
$$ {\Theta^\mu}_\nu \equiv -\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi^{(i)} )} \partial_\nu \phi^{(i)} + {\delta^\mu}_\nu \mathcal{L} \tag9$$
with the property that
$$ (\partial_\mu {\Theta^\mu}_\nu)\delta x^\nu + {\Theta^\mu}_\nu \partial_\mu (\delta x^\nu) + \partial_\mu F^\mu = 0$$
$$\Rightarrow (\partial_\mu {\Theta^\mu}_\nu)\delta x^\nu + \partial_\mu F^\mu = 0 \,. \tag{10}$$
Note we have used that $\delta x^\nu$ is a constant shift in space-time coordinates.
This quantity $ {\Theta^\mu}_\nu $ is a possible candidate for the stress-energy tensor ${T^\mu}_\nu $. However, recall that an important feature of the latter is that it is symmetric. We can use the flexibility in the choice of the auxiliary field $F^\mu$ to construct the symmetric tensor we require. Pick the following field
$$ F^\mu \equiv {\Theta_\nu}^\mu \delta x^\nu \tag{11} $$
and define the stress-energy tensor by
$$ T^{\mu\nu} \equiv \Theta^{\mu\nu} + \Theta^{\nu\mu} \,.\tag{12} $$
Eqn.(10) now implies that
$$ \boxed{\partial_\mu T^{\mu\nu} = 0} \tag{13}$$
NOTE/CAUTION!
For eqn.(11) to reflect a valid choice, one needs to assume proper asymptotic fall-off conditions for the fields, their derivatives and the Lagrangian. This is because the constant space-time shift $\delta x^\mu$ does not vanish at infinity, so that $\Theta^{\mu\nu}$ should compensate for this.
You considered a passive transformation and that did not work out well because of very good reasons. The stress-energy tensor of a system is a physical quantity that remains conserved if there is a symmetry of the system with respect to space-time translations (recall from mechanics that time translation symmetry relates to energy conservation and space translation symmetry yields momentum conservation). Therefore, you need to physically transform the field configuration only. The transformed field configuration should relate to the original one by a space-time translation. This is different from keeping the actual field configuration fixed and transforming your coordinate system and observing how the fixed physical fields appear from your new coordinate system. That is not what you want to do. The physical principle is that your fields have a certain symmetry and therefore you necessarily need to transform your fields to probe the nature of the symmetry. Hence, active transformations are required.
These notes are a great compilation of the fundamental physical principles required to understand the stress-momentum tensor.
Best Answer
This is a clever method used to derive Noether's current for any global symmetry; for the translational symmetry, it produces the stress-energy tensor.
We have to consider a local transformation because the variation of the action, $\delta S$, vanishes for the global transformation because the global transformation is by definition a symmetry: $$\delta S = 0$$ This value of $\delta S$ would follow "tautologically" and we couldn't deduce anything new out of it.
It follows that if we "generalize" the symmetry transformation rules and make them $x$-dependent, i.e. the transformation will be specified by $\delta a(x)$ (and $a$ is a vector for translations, but may be scalar for other symmetries), then $\delta S$ will be nonzero but it will inevitably depend on derivatives of $a$ only; for constant choices of $a$, we must get zero (because it's the global symmetry). For actions that only depend on first derivatives of the fields, the variation of the action will inevitably have the form $$ S = \int (\partial_\mu a) j^\mu d^d x$$ where $j^\mu$ is some particular function of the fields or other degrees of freedom (and their derivatives). Note that this form is inevitable: $\delta S$ has to be linear in $a$ and/or its derivatives, but it must vanish for $a={\rm const}$ so there can't be any terms of the form $\int ab\,\,d^dx$, i.e. terms proportional to undifferentiated $a$. There are also no higher-derivative terms if the action didn't have higher derivatives of fields to start with.
Now, the argument that $j^\mu$ is a current is simple. When equations of motion are satisfied, $\delta S =0$ for any variation of fields, whether it's a symmetry or not. In particular, $\delta S = 0$ holds for the "generalized" or "localized" global symmetry given by $a(x)$ which is no longer an exact symmetry, so $\delta S$ is the nonzero expression above. But by integration by parts, $\delta S = 0$ means $$ 0 = \int a(x)\cdot \partial_\mu j^\mu(x)\,\, d^d x$$ which vanishes iff $\partial_\mu j^\mu$ vanishes at each point. This proves that $j^\mu$ obtained in this way is a conserved current; its integral has to be a conserved quantity. You could ask why someone invented this method. He or she invented it because he or she was creative and smart. What's important for everyone else is to check the arguments above and see that one may derive a conserved current in this way. The original author of the method was able to "see" the whole argument in his or her head.
(I say "her" as well to pay some tribute to Noether who didn't quite invent this elegant method – her papers were messy – but she invented the whole relationship between symmetries and conservation laws.)