Geometric intuition of tensor product

You want to stay concrete, so let's let $V$ be a two-dimensional real vector space and $W = \operatorname{Hom}(V,\mathbb{R})$. Then $V = T^1_0(V)$ and $W = T^0_1(V)$, so for any $v\in V$ and $w\in W$, $v\otimes w\in T^1_1(V)$.

Each of $v,w$ has two components; $v = v^1e_1 + v^2e_2$ and $w = w_1e^1 + w_2e^2$. Here $e_1,e_2$ is a basis for $V$ and $e^1,e^2$ is the dual basis in $V^*$.

The components of $v\otimes w$ are all the components of $v$ times all the components of $w$: $$(v\otimes w)^i_j = v^iw_j.$$

To see this, observe that $(v\otimes w)(\theta, x)=v(\theta)w(x),$ so that $(v\otimes w)(e^i, e_j) = v(e^i)w(e_j).$

More generally, if $A = A^{i_1\cdots i_p}_{j_1\cdots j_q}\in T^p_q$ and $B = B^{k_1\cdots k_r}_{l_1\cdots l_s}\in T^r_s$, then

$$(A\otimes B)^{i_1\cdots i_pk_1\cdots k_r}_{j_1\cdots j_q l_1\cdots l_s} = A^{i_1\cdots i_p}_{j_1\cdots j_q}B^{k_1\cdots k_r}_{l_1\cdots l_s}.$$

Note that our example is from the derived tensor algebra over a two-dimensional vector space, $T^p_q(V) = (V^*)^{\otimes p}V^{\otimes q}.$ Hopefully this helps build your intuition about the case where $V$ and $W$ are two vector spaces of potentially different dimension.

The way I like to think about this is that if we have two vector spaces $V$ and $W$, their tensor product $V \otimes W$ intuitively is the idea "for each vector $v\in V$, attach to it the entire vector space $W$." Notice however that since we have $V\otimes W \cong W \otimes V$ we can rephrase this as the idea "for every $w \in W$ attach to it the entire vector space $V$." What distinguishes $V\times W$ from $V\otimes W$ is that the tensor space is essentially a linearized version of $V\times W$. To obtain $V\otimes W$ we start with the product space $V\times W$ and consider the free abelian group over it, denoted by $\mathcal{F}(V\times W)$. Given this larger space, that admits scalar products of ordered pairs, we set up equivalence relations of the form $(av,w) \sim a(v,w) \sim (v, aw)$ with $v\in V, w \in W$ and $a \in \mathbb{F}$ (a field), so that we identify points in the product space that yield multilinear relationships in the quotient space generated by these relationships. If $\mathcal{S}$ is the subspace that is spanned by these equivalence relations, we define $V\otimes W = \mathcal{F}(V\times W)/\mathcal{S}$.

Personally, I like to understand the tensor product in terms of multilinear maps and differential forms since this further makes the notion of tensor product more intuitive for me (and this is typically why tensor products are used in physics/applied math). For instance if we take an $n$-dimensional real vector space $V$, we can consider the collection of multilinear maps of the form

$$ F: \underbrace{V \otimes \ldots \otimes V}_{k \; times} \to \mathbb{R} $$

where $F \in V^* \otimes \ldots \otimes V^*$, and this tensor space has basis vectors $dx^{i_1} \otimes \ldots \otimes dx^{i_k}$ and $\{i_1, \ldots, i_k\} \subseteq \{1, \ldots, n\}$. Notice now that given $k$ vectors $v_1, \ldots, v_k \in V$ we have that multilinear tensor functional $dx^{i_1} \otimes \ldots \otimes dx^{i_k}$ acts on the ordered pair $(v_1, \ldots, v_k)$ as

$$ dx^{i_1} \otimes \ldots \otimes dx^{i_k}(v_1, \ldots, v_k) \;\; =\;\; dx^{i_1}(v_1) \ldots dx^{i_k}(v_k). $$

Where we have that $dx^{i_j}(v_j)$ takes the vector $v_j$ and picks out its $i_j$-th component. Extending this notion to differential forms where instead we would have basis elements $dx^{i_1} \wedge \ldots \wedge dx^{i_k}$ (where antisymmetrization is taken into account), we would have that this basis form would take a set of $k$ vectors and in some sense "measure" how much these vectors overlap with the subspace generated by the basis $\{x_{i_1}, \ldots, x_{i_k}\}$.

The difference between the ordered pair $(v,w)$ of vectors and the tensor product $v\otimes w$ of vectors is that for a scalar $c\not\in\{0,1\}$, the pair $(cv,\;w/c)$ is different from the pair $(v,w)$, but the tensor product $(cv)\otimes(w/c)$ is the same as the tensor product $v\otimes w$.