Dual vector space is the set of all linear functionals defined on a given vector space. The vector space and dual vector space is isomorphic and hence have the same dimension. A rank $(k,l)$ tensor is a multilinear functional defined on the product space consisting of $k$ dual spaces and $l$ vector spaces. Hence analogues to above,the set of all $(k,l)$ tensors should be isomorphic to the product space consisting of $k$ dual vector spaces and $l$ vector spaces, hence should have a dimension $4k+4l$, assuming we are considering the spacetime. But set of all $(k,l)$ rank tensors have $4^{k+l}$ basis vectors defined using tensor product hence have the dimension $4^{k+l}$. Why is that set of all tensors of rank $(k,l)$ is not isomorphic to the product space? How can we interpret these extra degree of freedom.
Special Relativity – Dimension of a Vector Space of All Tensors of Rank $(k,l)$ in 4D
linear algebraspacetimespacetime-dimensionsspecial-relativitytensor-calculus
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A $(p,q)$ tensor field $T$ on a smooth manifold $M$ is defined as a multilinear map: $$T : \underbrace{\Omega(M)\times\dots\times\Omega(M)}_{p\ times} \times\underbrace{\mathfrak{X}(M)\times\dots\times\mathfrak{X}(M)}_{q\ times} \to C^{\infty}(M)$$ where $\Omega(M)$ is the set of all covector fields on $M$ and $\mathfrak{X}(M)$ is the set of all vector fields on $M$.
Unpacking the above, $T$ sends $p$ covector fields $H_1,\ldots,H_p$ and $q$ vector fields $X_1,\ldots,X_q$ to some $C^{\infty}$ function $f$ on $M$. i.e., $T(H_1,\ldots,H_p,X_1,\ldots,X_q)$ is a $C^{\infty}$ function on $M$. Notationally, its effect for a point $x\in M$ is $$T_x(H_1(x),\ldots,H_p(x),X_1(x),\ldots,X_q(x))=f(x)$$ where $H_i(x)$ and $X_i(x)$ are respectively covectors and vectors in cotangent and tangent spaces at $x$. $T_x$ is defined locally at $x$ as: $$T_x:\underbrace{T^*_xM\times\dots\times T^*_xM}_{p\ times} \times\underbrace{T_xM\times\dots\times T_xM}_{q\ times} \to F$$
Thus you can see how the tensor field $T$ selects for each point $x$ a locally defined tensor acting on cotangent and tangent spaces at $x$.
To address what you wrote in the question: "a tensor is defined as a linear multilinear map on a set of vector spaces and/or dual vector spaces to a field..." - you hopefully see why this isn't accurate. Similarly, "tensor field is defined as a linear multilinear map on a set of tangent vector spaces and/or dual tangent vector spaces to a field" isn't accurate either.
Let $H$ be a (separable) Hilbert space, such as we have in quantum mechanics. You can say that its inner product is a map $H\times H\to\mathbb{C}, (\lvert v\rangle,\lvert w\rangle)\mapsto \langle v\vert w\rangle$, and for $\lvert v\rangle\in H$ we can define $\langle v\vert \in H^\ast$ by the map $H\to \mathbb{C}, \lvert w\rangle\mapsto \langle v \vert w\rangle$. This is how the idea of dual spaces arises naturally for inner product spaces - the inner product gives you a natural map $H\to H^\ast, \lvert v\rangle \to \langle v\rvert$ from the space to its dual and the Riesz representation theorem establishes that this map is an (anti-)isomorphism.
Therefore, it is mathematically completely equivalent to view $\langle v\vert w\rangle$ either as the inner product of two vectors in $H$, or as the action of an element of $H^\ast$ on $H$. There is rarely any "need" to consider a particular of these viewpoints - since they are equivalent, you can rephrase every statement about the inner product as a statement about the dual and vice versa - but some things might be easier to express.
Note that the dual space and the map from kets to bras is completely fixed by $H$ and its inner product - there is no choice here, there isn't any additional information needed to talk about the dual space, so it's a bit strange to ask whether we "need" the dual space - it's just something that exists.
The notion of the dual becomes more subtle and more relevant as something distinct from the inner product when you try to formalize what certain objects like the "eigenstates" $\lvert x_0\rangle$ of the position operator are. They aren't elements of the Hilbert space, which is easiest to see when people write them in the position basis as $f(x) = \delta(x - x_0)$ - this isn't even a proper function, let alone an element of the Hilbert space of square-integrable functions $L^2$. However, the $\delta$-function is a tempered distribution $S^\ast$, which is the dual space to the space of Schwartz functions $S$. That is, if you want to be careful about it, $\langle x\rvert$ exists as a bra on the space of "Schwartz kets", but not as a ket, nor as a bra on the full Hilbert space. The sequence of spaces $S\subset L^2 \subset S^\ast$ is known as a Gel'fand triple and leads one to the notion of a rigged Hilbert space.
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Correct (assuming you mean, as I think you really do, the Cartesian product rather than tensor product).
This is wrong! We want the dimension of the space of multilinear functionals, not the dimension of the domain of these multilinear functionals (which is what you’re calculating). We’re looking for all possible multilinear maps on a large domain space, which is why we get a potentially larger space.
I would start off with the following comment: it is usually good to reduce a multilinear algebra problem to a linear algebra one. Suppose we have three vector spaces, $V_1,V_2,W$, over a fixed field $\Bbb{F}$. Let $\text{Hom}^2(V_1\times V_2;W)$ be the vector space of bilinear maps $V_1\times V_2\to W$. There is a canonical isomorphism \begin{align} \Phi:\text{Hom}^2(V_1\times V_2;W)\to \text{Hom}(V_1,\text{Hom}(V_2,W)) \end{align} defined simply as $\Phi(T)= [v_1\mapsto T(v_1,\cdot)]$, i.e given a bilinear $T$, we define $\Phi(T)$ to be the linear map which assigns to each $v_1\in V_1$, the linear map $T(v_1,\cdot)= [v_2\mapsto T(v_1,v_2)]$. I leave it to you to verify that $\Phi$ is a linear map, and that it is invertible with inverse \begin{align} \Phi^{-1}:\text{Hom}(V_1,\text{Hom}(V_2,W))\to \text{Hom}^2(V_1\times V_2;W) \end{align} given by $\left(\Phi^{-1}(S)\right)(v_1,v_2):=\left(S(v_1)\right)(v_2)$.
To understand this, you just need to keep your head straight about the order of evaluation, and the idea of ‘freezing one variable’, or as is also known, currying. Therefore, the above argument shows that these two vector spaces are isomorphic, and hence have the same dimension: \begin{align} \dim \text{Hom}^2(V_1\times V_2;W)&=\dim \text{Hom}(V_1,\text{Hom}(V_2,W))\\ &=\dim V_1\cdot \dim \text{Hom}(V_2,W)\\ &=\dim V_1\cdot \dim V_2\cdot \dim W, \end{align} because the dimension of the space of linear maps between two vector spaces is the product of dimensions (the pedestrian way of thinking is that an $m\times n$ matrix has $m\cdot n$ entries which can be filled, not $m+n$). Again, note carefully that this is different from the dimension of the domain, $\dim(V_1\times V_2)=\dim V_1+\dim V_2$.
Extending this idea using induction (take $V_1=\dots= V_k=V^*$ and $V_{k+1}=\cdots = V_{k+l}=V$ and $W=\Bbb{F}$ the underlying field), you see that the $(k+l)$-fold multilinear maps form a vector space of dimension $(\dim V)^{k+l}$ (assuming $\dim V$ is finite of course, so that $\dim V=\dim V^*$).