Note. All vector spaces are of finite dimension. The question concerns the coherence between the tensor product of vector spaces and the tensor product between vectors. But first let's introduce some definitions.
Let $V_1,\dots, V_p$ be vector spaces over a field $\mathbb{K}$. For us $$V_1\otimes\cdots\otimes V_p=\text{Multilinear}(V_1^{*},\dots ,V_p^{*};\mathbb{K}),$$ that is the set of the $p-$ linear application from $V_1^{*}\times\cdots\times V_p^{*}$ in $\mathbb{K}$.
Consider the $p-$linear application $$F\colon V_1\times \cdots \times V_p \to V_1\otimes\cdots\otimes V_p$$ defined as $$F(v_1,\dots, v_p)(\varphi^1,\dots, \varphi^p)=\varphi^1(v_1)\cdots\varphi^p(v_p),$$ for all $v_1\in V_1,\dots, v_p\in V_p$ and $\varphi^1\in V_1^{*},\dots, \varphi^p\in V_p^{*}$.
The elements of $V_1\otimes\cdots\otimes V_p$ are called tensors, while the elements of the type $F(v_1,\dots, v_p)$ are called decomposable tensors.
Let $V$ be a vector space over a field $\mathbb{K}$. We can consider the following vectorial space $$T_0^0(V)=T_0(V)=T^0(V)=\mathbb{K};$$ $$T^1(V)=T_0^1(V)=V;$$ $$T_1(V)=T_1^0(V)=V^*;$$ $$T^p(V)=T_0^p(V)=V\otimes\cdots\otimes V\quad\text{(p factors)}$$ $$T_q(V)=T_q^0(V)=V^*\otimes\cdots\otimes V^*\quad\text{(q factors)}$$ $$T_q^p(V)=T^p(V)\otimes T_q(V)$$ $$T(V)=\bigoplus_{p,q\ge 0}T_p^q(V)$$
If $\{v_1,\dots, v_n\}$ is a basis of $V$ and $\{v^1,\dots. v^n\}$ the dual basis of $V^*$. Then the basis of $T_p^q(V)$ is $$v_I=v_{i_1}\otimes\cdots\otimes v_{i_p}\otimes v^{i_{p+1}}\otimes\cdots\otimes v^{i_{p+q}}.$$
Remark Since $T_q^p(V)$ is the set of the $(p+q)-$ linear applications from $(V^*)^p\times V^q$ to $\mathbb{K}$, the action of the decomposable tensors is given by $$F(u_1,\dots, u_p,\omega^1,\dots, \omega^p)(\eta^1,\dots, \eta^p, v_1,\dots, v_q)=\eta^1(u_1)\cdots\eta^p(u_p)\cdot\omega^1(v_1)\cdots \omega^q(v_q),$$ where $u_1,\dots, u_p, v_1,\dots, v_q\in V$ and $\omega^1,\dots, \omega^q,\eta^1\dots, \eta^p\in V^{*}$.
We want to define a product on $T(V)$ and here the question will soon be triggered.
Definition.(Tensor Product) If $\alpha\in T_{q_1}^{p_1}(V)$ and $\beta\in T_{q_2}^{p_2}(V)$, then we define $\alpha\otimes \beta\in T_{q_1+q_2}^{p_1+p_2}(V)$ setting
$$\alpha\otimes \beta (\eta^1,\dots, \eta^{p_1+p_2}, v_1,\dots, v_{q_1+q_2})=\alpha(\eta^1,\dots \eta^{p_1},\dots, v_1,\dots, v_{q_1})\beta(\eta^{p_1+1},\dots, \eta^{p_1+p_2}, v_{q_1+1},\dots, v_{q_1+q_2})$$ for all $\eta^1,\dots, \eta^{p_1+p_2}\in V^*$ and for all $v_1,\dots, v_{q_1+q_2}\in V$.
Question Why the tensor product of elements of $V$ or $V^*$ is exactly the decomposable tensor $F(\alpha, \beta)$? That is, is this product well defined when $\alpha$ and $\beta$ are vectors?
So, in this case, we start taking $\alpha\in T^1(V)$ and $\beta\in T^1(V)$ and we must prove that $\alpha\otimes \beta =F(\alpha, \beta)$. In this case $\alpha\otimes\beta\colon V^{*}\times V^{*}\to \mathbb{K}$ and $F(\alpha, \beta)\colon V^*\times V^*\to\mathbb{K}$.
I am aware that the question could be trivial, but I have not been able to formally show it for days, I would be very grateful if you would help me.
Following your definitions: let $\varphi_1,\varphi_2\in V^*$, then $$ [F(\alpha,\beta)](\varphi_1,\varphi_2)=\varphi_1(\alpha)\varphi_2(\beta) $$ and $$ (\alpha\otimes\beta)(\varphi_1,\varphi_2)=\varphi_1(\alpha)\varphi_2(\beta). $$ This holds for any $\varphi_1,\varphi_2\in V^*$, hence the maps are identical.
I have to note that I find the definition for the tensor product as the space of multi-linear mappings on the duals a bit cumbersome. It makes more sense to define it using the universal property of the tensor product.