Let $V$ be a vector space over a field $k$, let $w \in \otimes^l V$ be a tensors. We call $w$ a simple tensor if it can be written as $$ w=w_1 \otimes w_2 \otimes \ldots \otimes w_l, $$ where $w_i \in V$ for $i=1,\dots, l$. Simple tensors are also called rank $1$ tensors, and a tensor $w$ is called rank $r$ tensor if $r$ is the minimum number of terms in a presentation of $w$ as a sum of simple tensors.
With any tensor $w$ we can associate a finite sequence of linear maps $$ L_i : \otimes^i V^* \to \otimes^{l-i} V, $$ where $1 \leq i \leq l-1$, $V^*$ is the dual space, and the maps $L_i$ acts on $\otimes^i V^*$ by evaluating tensor product of vectors on a tensor product of forms.
Is it true that if $\operatorname{rk}(L_i)=1$ for $1 \leq i \leq l-1$ then tensor $w$ is simple? In other direction it is clearly true. Thus, it could provide a criterion for a tensor to be simple in terms of rank of linear maps.
If the answer to previous question is positive, is it true for higher rank tensors?
The answer for your first question is yes. The answer for your second is no.
The next proposition is well known. It was proved here.
Proposition: Let $V,W$ be vector spaces over $k$. Let $w=\sum_{i=1}^ra_i\otimes b_i=\sum_{i=1}^sv_i\otimes w_i\in V\otimes W$.
Answer for problem 1:
Now, let $w\in\bigotimes^lV$. Thus we can consider $w\in V\otimes W$, where $W=\bigotimes^{l-1}V$.
Assume each $L_i$ has rank $1$. Let $w=\sum_{i=1}^ra_i\otimes b_i$, where $a_i\in V$ and $b_i\in W$ and $\{a_1,\ldots, a_r\}$, $\{b_1,\ldots, b_r\}$ are linear independent. We can find $f_i\in \bigotimes^{l-1}V^*$, for $1\leq i\leq r$, such that $f_i(b_j)=\delta_{ij}$, for $1\leq i,j\leq r$.
Thus, $Id\otimes f_i(w)=a_i$, for $1\leq i\leq r$. Thus, one of your maps, $L_{l-1}$, has rank $r$. Therefore, $r=1$ and $w=a_1\otimes b_1$, where $a_1\in V$ and $b_1\in \bigotimes^{l-1}V$. By induction on $l$, we can prove that $b_1=v_2\otimes\ldots\otimes v_l$ and we are done.
Answer for problem 2:
Now, let us prove that this idea does not work for rank 2. Now, let $a,b\in V$ be linear independent. Consider $w=a\otimes a\otimes a+a\otimes b\otimes b+b\otimes a\otimes b$. Notice that any map $L_i$ has rank 2. Please check it.
Now, assume $w= c\otimes d\otimes e+f\otimes g\otimes h$, for some $c,d,e,f,g,h\in V$.
Notice that $w$ as a tensor in $V\otimes W$, where $W=\bigotimes^2V$, has tensor rank 2, because $w=a\otimes(a\otimes a+b\otimes b)+b\otimes (a\otimes b)$ and by item 3 of proposition above.
Thus, $\{c,f\}$ must be linear independent. Analogously, we can prove that $\{d,g\}$ is linear independent and $\{e,h\}$ is too. Using this proposition again we have $\text{span}\{a,b\}=\text{span}\{c,f\}=\text{span}\{d,g\}=\text{span}\{e,h\}$.
Let $\delta_a,\delta_b\in V^*$ be such that $\delta_a(a)= \delta_b(b)=1$ and $\delta_a(b)=\delta_b(a)=0$.
Thus, $Id\otimes Id\otimes \delta_a(w)=a\otimes a= c\otimes d\delta_a(e)+f\otimes g \delta_a(h)$.
Since $\{c,f\}$ and $\{d,g\}$ are linear independent then, by item 3 of the proposition, we must have or $\delta_a(e)=0$ or $\delta_a(h)=0$. Assume $\delta_a(e)\neq0$ and $\delta_a(h)=0$.
Now remind that $e=\delta_a(e)a+\delta_b(e)b$ and $h=\delta_b(h)b$, because $\text{span}\{a,b\}=\text{span}\{e,h\}$.
Thus, $c\otimes d\otimes e=c\otimes d\otimes (\delta_a(e)a+\delta_b(e)b)=a\otimes a\otimes a+a\otimes a\otimes \dfrac{\delta_b(e)}{\delta_a(e)}b$.
Therefore $a\otimes b\otimes b+b\otimes a\otimes b=a\otimes a\otimes \dfrac{\delta_b(e)}{\delta_a(e)}b+f\otimes g\otimes \delta_b(h)b$
Thus, $m=a\otimes (b-\dfrac{\delta_b(e)}{\delta_a(e)}a)\otimes b+b\otimes a\otimes b=f\otimes g\otimes \delta_b(h)b$.
Finally, $Id\otimes Id\otimes \delta_b(m)=a\otimes (b-\dfrac{\delta_b(e)}{\delta_a(e)}a)+b\otimes a=f\otimes g\delta_b(h)$, which is impossible. The left side of the equation has tensor rank 2 and the right side has tensor rank 1, by item 3 of proposition.