Let ${^ts}:E\to F$ be the transpose of $s:F\to E$. Show that $\text{Im}(s)\cap \ker (\,^ts)=\{0_E\}$. 
Let $s\in \mathcal{L}(F,E)$
$$\displaystyle F \overset{s}{\longrightarrow} E\overset{^ts}{\longrightarrow} F$$
I spent two days to show that :$$\text{Im}(s)\cap \ker (\,^ts)=\{0_E\}\qquad \tag{1}$$

I'm not sure that is right, I tried with specific matrix (3x3), and it works.
But when I want to provide a general proof, I struggle.
I used the Annihilator but no way to find a solution, may be this statement is wrong...?
I add more information about $E$ and $F$ regarding the comments
$F=\mathbb{R}^p$ and $E=\mathbb{R}^n$ with $p\le n$
Or $F=(\mathbb{R}^p,\langle\cdot,\cdot\rangle)$ and $E=(\mathbb{R}^n,\langle\cdot,\cdot\rangle)$
The matrix associated at $u$ is $M$ and $M\in \mathcal{M}_{n,p}(\mathbb{R})$
 A: It is  sufficient to show that 
$$
\ker s^t\subset(\text{Im } s)^\perp 
$$
because we can then conclude by:
$$
\ker s^t \cap \text{Im}\ s \subset (\text{Im}\ s)^\perp \cap \text{Im}\ s =\{0_E\}
$$

The claim $\ker s^t\subset(\text{Im } s)^\perp$ can be proven as follows:
\begin{align*}
e\in\ker s^t &\Rightarrow s^t(e)=0_F \\
&\Rightarrow  \forall f\in F,\ \langle f,s^t(e) \rangle_F=0 \\ &\Rightarrow  \forall f\in F,\ \langle s(f),e\rangle_E=0 \\ 
&\Rightarrow  \forall f\in F,\ s(f) \perp e \\
&\Rightarrow  e \in (\text{Im}\ s)^\perp
\end{align*}
A: In order that the transpose is a map $F\to E$ you need some assumptions: both vector spaces need to be equipped with a nondegenerate bilinear form, so that we can define isomorphisms $E\to E^*$ and $F\to F^*$ (the dual spaces), assuming finite dimensionality.
In particular, if the base field is $\mathbb{R}$, the two space might be inner product spaces.
If $\langle{\cdot},{\cdot}\rangle_E$ is the form on $E$, then for each $v\in E$, the map
$$
e_v\colon E\to K,\qquad e_v(x)=\langle v,x\rangle_E
$$
is an element of $E^*$. If $v\ne0$, then nondegeneracy implies there exists $x\in E$ with $\langle v,x\rangle_E\ne0$, so $e\colon E\to E^*$, $v\mapsto e_v$ is an isomorphism (it is clear from bilinearity that $e_v$ is linear, for every $v\in E$, and $e$ is linear as well).
Then the transpose ${}^{t\!}s$ is the dual map composed with these isomorphisms:
$$
{}^{t\!}s=e^{-1}\circ s^*\circ e
$$
(I denote by $e$ both maps $E\to E^*$ and $F\to F^*$, no confusion should arise).
If $w=s(v)$ and ${}^{t\!}s(w)=0$, then also $s^*\circ e_w=0$, which means $e_w\circ s=0$, that is,
$$
e_w(s(x))=0\quad\text{for all $x\in V$}
$$
hence
$$
\langle w,s(x)\rangle_F=0
$$
In particular, $\langle s(v),s(v)\rangle_F=0$, so by nondegeneracy, $s(v)=0$ and $w=0$.
A: If you represent a linear map by a matrix then the image of the map is the sub-space spanned of the columns of the matrix. And the kernel of the map is the sub-space that is perpendicular to the rows of the matrix (because the inner product of each row with any vector in the kernel must be zero).
So the image of $s$ is the sub-space that is spanned by the columns of the matrix representing $s$ (relative to specific bases in $E$ and $F$).
And the kernel of $^ts$ is the sub-space that is perpendicular to the sub-space spanned by the rows of the matrix representing $^ts$.
But the columns of the first matrix are the rows of the second matrix. The result follows.
A: Let $s:F\to E$ be a linear transformation from a vector space $F$ to another vector space $E$ over the base field $\mathbb{R}$.  Both $E$ and $F$ are equipped with nondegenerate symmetric bilinear forms $\langle\_,\_\rangle_E$ and $\langle\_,\_\rangle_F$, respectively.  Suppose further that $\langle\_,\_\rangle_E$ is positive-definite.  Write $s^\top:E\to F$ for the transpose of $s$.  We claim that
$$\text{im}(s)\cap\ker(s^\top)=\{0_E\}\,.$$
Recall that $s^\top$ is the unique linear transformation from $E$ to $F$ such that $$\big\langle s(x),y\big\rangle_E=\big\langle x,s^\top(y)\rangle_F$$
for all $x\in F$ and $y\in E$.  Suppose that $y\in\text{im}(s)\cap\ker(s^\top)$.   Then, $y\in\text{im}(s)$, so $y=s(x)$ for some $x\in F$.  Sinc $y\in\ker(s^\top)$, we get $$s^\top\big(s(x)\big)=s^\top(y)=0_F\,.$$
Therefore,
$$\Big\langle x,s^\top\big(s(x)\big)\Big\rangle_F=\big\langle x,0_F\big\rangle_F=0\,.$$
By the definition of $s^\top$, we have
$$\langle y,y\rangle_E=\big\langle s(x),s(x)\big\rangle_E=\Big\langle x,s^\top\big(s(x)\big)\Big\rangle_F=0\,.$$
Since $\langle\_,\_\rangle_E$ is positive-definite, we have $y=0_E$.  This proves the claim.
P.S.:  If the ground field is $\mathbb{C}$, we instead assume that $\langle\_,\_\rangle_E$ and $\langle\_,\_\rangle_F$ are nondegenerate Hermitian forms.  The transpose map has to be replaced by the Hermitian conjugate.  For the claim to hold, we need to assume further that $\langle\_,\_\rangle_E$ is positive-definite.

If the ground field $\mathbb{K}$ is a field of characteristic $0$, but there is no positive-definiteness assumption on $\langle\_,\_\rangle_E$ (which makes no sense when $\mathbb{K}$ is not a subfied of $\mathbb{R}$ anyhow), then the claim is not true.  For example, let $\mathbb{K}:=\mathbb{R}$, $E:=\mathbb{R}^2$, and $F:=\mathbb{R}^2$.  Then we set $e_1:=(1,0)$, $e_2:=(0,1)$.  Let $\langle\_,\_\rangle$ be the nondegenerate symmetric bilinear form on both $E$ and $F$ given by $$\langle a_1e_1+a_2e_2,b_1e_1+b_2e_2\rangle:=a_1b_1-a_2b_2$$
for all $a_1,a_2,b_1,b_2\in\mathbb{R}$.  Take $s:F\to E$ to be the linear map sending both $e_1$ and $e_2$ to $e_1+e_2$.  That is, $s$ is given by the matrix
$$\begin{bmatrix}1&1\\1&1\end{bmatrix}$$
with respect to the ordered basis $(e_1,e_2)$ of both $F$ and $E$.  Then, $s^\top:E\to F$ is given by the matrix
$$\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$$
with respect to the ordered basis $(e_1,e_2)$ of both $E$ and $F$.  This shows that $$\text{im}(s)=\text{span}_\mathbb{R}\{e_1+e_2\}=\ker(s^\top)\,.$$

The claim fails in positive characteristics.  If the ground field $\mathbb{K}$ is of a positive characteristric $p$, then take $E$ and $F$ to be both $\mathbb{K}^p$.  Fix a basis $\{e_1,e_2,\ldots,e_p\}$ of both $E$ and $F$.  Equip $E$ and $F$ with the nondegenerate symmetric bilinear form $\langle\_,\_\rangle$ given by
$$\left\langle\sum_{i=1}^p\,a_ie_i,\sum_{i=1}^p\,b_ie_i\right\rangle:=\sum_{i=1}^p\,a_ib_i$$
for all $a_1,a_2,\ldots,a_p,b_1,b_2,\ldots,b_p\in\mathbb{K}$.  If the map $s:F\to E$ is the $\mathbb{K}$-linear map sending $e_i\mapsto e_1+e_2+\ldots+e_p$ for all $i=1,2,\ldots,p$, then its transpose $s^\top:E\to F$ also sends $e_i\mapsto e_1+e_2+\ldots+e_p$ for all $i=1,2,\ldots,p$.  In particular, this shows that $$\text{span}_\mathbb{K}\{e_1+e_2+\ldots+e_p\}=\text{im}(s)\cap\ker(s^\top)\,.$$
