I think that your example does not work. Indeed, we have the following lemma.
Lemma (Uniform surjectivity implies product surjectivity)
Let $\varphi_i : \mathcal{F}_i \to \mathcal{G}_i$ be a collection of sheaf morphisms, with base space $X$. Suppose that there is a basis $\mathcal{B}$ of $X$ such that $\varphi_i(U):\mathcal{F}_i(U) \to \mathcal{G}_i(U)$ is surjective for all $U \in \mathcal{B}$. Then $\prod_i \mathcal{F}_i \to \prod_i \mathcal{G}_i$ is sheaf surjective.
Proof. Sheaf surjectivity is equivalent to surjectivity on stalks. Take $[(g_i,U)]$ be an element of the stalk $(\prod \mathcal{G}_i)_x$. Take $x \in V \subset U$ such that $V \in \mathcal{B}$. By hypothesis, there exist $[(f_i, V)]$ that are sent to $[(g_i |V, V)]$, and we are done.
Corollary. Your example does not work.
Indeed, if I well remember from complex analysis, your sheaf function is surjective on opens that are disks, and these constitute a basis.
Having in mind that we must force some disuniformity in $i$ with respect to the opens where we have a lifting, we build the following example.
For $U \subset X$, define $\mathbb{Z}_U$ as the sheafification of the presheaf $Z_U$ such that $Z_U(V) = \mathbb{Z}$ if $V \subset U$, and 0 otherwise. Restriction functions are idenitites when possible and zeros otherwise. To be honest, it is possible that $Z_U$ is already a sheaf, but we don't care since we are only interested in stalks and these are preserved by sheafification.
We have that $(\mathbb{Z}_U)_x = \mathbb{Z}$ if $x \in U$, and 0 otherwise, as a direct computation yields.
If $\mathbb{Z}_{U}(V) \neq 0$ and $V$ is connected, then $V \subset U$. Indeed, a section $s \in \mathbb{Z}_{U}(V)$ is given by a cover $\mathcal{V}_{i \in I}$ of $V$ and coherent sections $s_i \in \mathcal{V}_i$. Suppose $V \not \subset U$. Define $\Sigma$ as the poset of subsets $Q \subset I$ such that $s_i=0$ for all $i \in Q$. Then it is non empty, because there exist i s.t $V_i \not \subseteq U$, and evidently every ascending chain has an upper bound. We can henceforth apply Zorn lemma a find a maximal $J$. Suppose by contradiction that $J \neq I$. There exist $i \in I \setminus J$ and $j\in J$ such that $V_i \cap V_j \neq \emptyset$, otherwise they would disconnect $V$. $V_i$ must be contained in $U$, otherwise $s_i$ would be zero. But then $s_i = s_i | V_i \cap V_j = s_j | V_i \cap V_j = 0$.
Take $X=\mathbb{R}^k$, and $U_n =$ disk of radius $1/n$ around 0. Take the skyscraper sheaf $\mathcal{G}$ centered at zero with stalk $\mathbb{Z}$, i.e. $\mathcal{G}(U) = \mathbb{Z}$ if $0 \in U$ and 0 otherwise, with restriction given by identities when possible and zero otherwise. Define $\varphi_n: \mathbb{Z}_{U_n} \to \mathcal{G}$ in the following way. It is enough to define $\phi_n: Z_{U_n} \to \mathcal{G}$, and then $\varphi_n$ follwos by universal property of sheafification. If $V \subset U_n$, define $\phi_n(V) = 1_{\mathbb{Z}}$ and zero otherwise. It commutes with restrictions. Indeed, let $V \subset V'$ be subsets. If $V' \not \subset U_n$, the diagram starts with a zero. If $0 \not \in V$, the diagram ends with a zero. If $0 \in V \subset V' \subset U_n$, then the commuting diagram is made of all identities.
- Consider the sheaf morphism
$$\prod \varphi_n: \prod_{n \in \mathbb{N}} \mathbb{Z}_{U_n} \to \prod_{n \in \mathbb{N}} \mathcal{G}$$
Take the element $[(\{1\}_{n \in \mathbb{N}}, U_1)]$ in $( \prod_n\mathcal{G} )_0$, and suppose that there exist $[(\{f_n\}_{n \in \mathbb{N}},V)]$ that is sent to $[(\{1\}_{n \in \mathbb{N}}, U_1)]$. This means that, after further restricting to $V'$ if necessary, $f_n | V' = 1 $ for all $n \in \mathbb{N}$. Take $V''$ to be the connected component of $V'$ containing $0$. Recall that $V''$ is open. This imples in particular $\mathbb{Z}_{U_n}(V'') \not 0$ for all $ n \in \mathbb{N}$, i.e. $V'' \subset \bigcap_n U_n = \{0\}$; which is a contradiction.
On the other hand, the functions $\varphi_n$ are surjective, because they are surjective on stalks, and we are done.
There is something kind of subtle here that I didn't realize when I made my comment, so I'll flesh it out right now for my own sake. When defining the image of a morphism $f:A\to B$ in an abelian category, we usually say $\mathrm{im}(f)$ is defined to be either $\mathrm{coker}(\mathrm{ker}(f))$ or $\mathrm{ker}(\mathrm{coker}(f))$ and we leave it to the exercises to show that these two definitions are the same. The thing is that when we say $\mathrm{ker}(f)$ for example, we are not referring to an object but instead another morphism $K\to A$ with a universal property. Similarly $\mathrm{coker}(f)$ is really a morphism $B\to C.$ So then what do we mean when we say $\mathrm{coker}(\mathrm{ker}(f))\cong \mathrm{ker}(\mathrm{coker}(f))?$ Since one is a morphism out of $A$ and the other is a morphism into $B,$ this isn't immediately clear. The answer is that the objects in question are isomorphic and the morphisms are the ones induced by the universal property of the other.
For example, denote the object $\mathrm{coker}(\mathrm{ker}(f))$ by $C$ and the object $\mathrm{ker}(\mathrm{coker}(f))$ by $K.$ Then $A\to C$ is the morphism attached to the object $C.$ By the universal property of the cokernal there is a unique map $C\to B$ since $\mathrm{\ker}(f)\to A\to B$ is zero. This map $C\to B$ is isomorphic to $K\to B.$ Similarly we can get a map $A\to K$ which is isomorphic to $A\to C.$
Okay now that all that is out of the way, here's what to do. Since $\mathrm{ker}(f)=0,$ the identity $1_A:A\to A$ satisfies the universal property of $\mathrm{coker}(\mathrm{ker}(f))=\mathrm{coker(0\to A)}.$ Since $0\to A\to B$ is zero, by the universal property of the cokernal, there is a unique map $A\to B$ making the diagram commute. But this is just $f:A\to B$ since our cokernal is just the identity. As stated in the middle paragraph, after composing with an isomorphism this is exactly the map $\mathrm{ker}(\mathrm{coker}(f))\to B.$ Hence $f=\mathrm{im}(f).$
Best Answer
Well, the expression $\widetilde{T} \colon F/\ker(T) \to \overline{\operatorname{im}(T)}$ immediately suggests something: the image of $\widetilde{T}$ is obviously $\operatorname{im}(T)$, so $\widetilde{T}$ can't be surjective unless $\operatorname{im}(T)=\overline{\operatorname{im}(T)}$. So just take any example of a bounded linear map whose image is not closed, and $\widetilde{T}$ will not be an isomorphism.