Presumably the solution you have in mind is this. There is a natural homomorphism $\newcommand{\E}{\mathcal{E}}\newcommand{\O}{\mathcal{O}}\newcommand{\Hom}{\mathcal{H}om}
\E\to(\E^\vee)^\vee$. It suffices to check that this homomorphism is an isomorphism at every stalk.
That's the right approach, but there's no need to go all the way to stalks. It suffices to show that on any open set $U\subseteq X$ where $\E$ is actually free, the natural map is an isomorphism. Since $\E|_U\cong \O_U^n$, it suffices to check that the natural map $\O^n\to \Hom(\Hom(\O^n,\O),\O)$ is an isomorphism. This is easy to check, but requires you to unravel the map.
Remark: Note that to show two sheaves on $X$ are isomorphic, it is not enough to find an open cover $X$ and isomorphisms between the two sheaves on each open set. The reason is that the isomorphisms may not agree the intersections of the open sets in the cover. Naturality of the map plays a very important role here: it ensures that the isomorphisms on the open cover will glue. To put it another way, we first constructed the morphism $\E\to (\E^\vee)^\vee$, and then checked that it is an isomorphism on an open cover. I want to stress that it is not enough to just check that $\O^n$ and $\Hom(\Hom(\O^n,\O),\O)$ are isomorphic, you must check that the specific map $s\mapsto (\phi\mapsto \phi(s))$ is an isomorphism.
This same remark holds if you want to use the stalks approach. You must first construct the global map, and then verify that it induces isomorphisms on stalks. Just showing that the stalks are isomorphic is not enough. If it were, any two locally free sheaves of the same rank would be isomorphic.
I gave you a bad hint in the comments because I had not thought about this in a while. Let me try to make up with it by giving you a (hopefully enlightning) solution. The key issue in your attempted solution is that you're not really engaging with the method suggested by Hartshorne: we can define a filtration of $\bigwedge^r\mathcal{F}$ which has successive subquotients $\bigwedge^p \mathcal{F}'\otimes\bigwedge^{r-p}\mathcal{F}''$. We define this filtration locally, and then check that it's "independent enough" to patch together globally.
We're going to use the same strategy as part (c): we're going to define a filtration of $\bigwedge^r \mathcal{F}$,
$$\bigwedge^r\mathcal{F} = F^0\supseteq F^1\supseteq\cdots\supseteq F^r \supseteq F^{r+1}=0 $$
which has successive quotients $F^p/F^{p+1}=\bigwedge^p\mathcal{F}' \otimes \bigwedge^{r-p} \mathcal{F}''$. Considering $r=\operatorname{rank} \mathcal{F}$, this will imply the result, as the filtration will reduce to $\bigwedge^r \mathcal{F} \cong \bigwedge^{rk \mathcal{F}'}\mathcal{F}'\otimes \bigwedge^{rk \mathcal{F}''}\mathcal{F}''$.
The way we construct this filtration is that we set it up in a basis-independent way on open subsets $U\subset X$ where each of $\mathcal{F}',\mathcal{F},\mathcal{F}''$ are free, and then check that this means it glues in to a filtration of the global sheaves. Fix an arbitrary such $U$, and choose any splitting $\mathcal{F}|_U\cong \mathcal{F}'|_U\oplus \mathcal{F}|_U''$. This gives us $$\bigwedge^r\mathcal{F}|_U \cong \bigoplus_{i=0}^r \left(\bigwedge^i \mathcal{F}'|_U\right)\otimes \left(\bigwedge^{r-i} \mathcal{F}''|_U\right).$$
Now we construct the filtration by induction - the idea of our strategy is to set $F^i$ to be those wedge products which have at least $i$ entries coming from a basis of $\mathcal{F}'|_U$. To do this, set $F^{r+1}=0$, and assume we've picked $F^{j+1},\cdots,F^{r+1}$ satisfying the requested properties. To construct $F^j$, consider the image of
$$\varphi: \bigwedge^j\mathcal{F}'|_U \otimes \bigwedge^{r-j} \mathcal{F}''|_U \to \left(\bigwedge^r \mathcal{F}|_U\right)/F^{j+1}.$$
I claim the preimage of this under the projection $\pi:\bigwedge^r\mathcal{F}|_U\to \left(\bigwedge^r\mathcal{F}|_U\right)/F^{j+1}$ is independent of the chosen splitting: pick a basis $x_1,\cdots,x_s$ for $\mathcal{F}'|_U$ and a basis $y_1,\cdots,y_t$ for $\mathcal{F}''|_U$, where we identify each with their image inside $\mathcal{F}|_U$. The image of $\varphi$ is the span of the collection $$x_{p_1}\wedge \cdots\wedge x_{p_j}\wedge y_{q_1}\wedge\cdots\wedge y_{q_{r-j}}$$ as the $p$ range over $1,\cdots,s$ and the $q$ range over $1,\cdots,t$. Then for any other basis $y_1+c_1,\cdots,y_t+c_t$ with $c_i\in \mathcal{F}'|_U$, the image is the span of the collection $$x_{p_1}\wedge \cdots\wedge x_{p_j}\wedge (y_{q_1}+c_{q_1})\wedge\cdots\wedge (y_{q_{r-j}}+c_{q_{r-j}}).$$ But expanding the wedge product in this sum, we see that it differs from a vector in our original collection by wedge products which have at least $j+1$ entries which are selected from the $x$s, and thus this difference is in $F^{j+1}$, so our claim about basis-independence is proven, and we can set $F^j$ to be this preimage.
This means our filtration $F^j$ is basis-independent, and in particular, it's compatible with restriction maps between open subsets where $\mathcal{F},\mathcal{F}',\mathcal{F}''$ are all free since any basis of the sheaves on the larger set is still a basis on the smaller set. Since any scheme is covered by such opens, this means we can patch our filtration together to an honest filtration of sheaves with the specified quotients globally, and we're done.
Best Answer
If $\mathscr{F} \cong \mathcal{O}_X^{\oplus n}$ is free, then the presheaf $\mathscr{P} \cong \mathcal{O}_X^{\oplus n^r}$ is also free, and hence is already a sheaf. For $U \subset X$ open, $$T^r(\mathscr{F})|_{U} = T^r(\mathscr{F}|_{U}),$$ so the result follows.
EDIT: I'll elaborate a bit on the isomorphism $\mathscr{P} \cong \mathcal{O}_X^{\oplus n^r}$. Let $e_1, \dots, e_n$ be a basis for $\mathscr{F}(X)$, so that $e_{i_1} \otimes \cdots \otimes e_{i_r}$ form a basis for $\mathscr{P}(X)$. These global sections define an isomorphism $\mathcal{O}_X^{\oplus n^r} \to \mathscr{P}$, because for $U \subset X$, $$e_{i_1}|_U \otimes \cdots \otimes e_{i_r}|_{U} = (e_{i_1} \otimes \cdots \otimes e_{i_r})|_{U}$$ form a basis for $\mathscr{P}(U)$. Indeed, $e_i|_{U}$ form a basis for $\mathscr{F}(U)$.