Let's first look at a very special type of tensor, namely the $(0,1)$ tensor. What is it? Well, it is the tensor product of $0$ copies of members of $V$ and one copy of members of $V^*$. That is, it is a member of $V^*$.
But what is a member of $V^*$? Well, by the very definition of $V^*$ is is a linear function $\phi:V\to K$. Let's write this explicitly:
$$T^0_1V = V^* = \{\phi:V\to K\mid\phi \text{ is linear}\}$$
You see, already at this point, where we didn't even use a tensor product, we get a $V^*$ on one side, and a $V$ on the other, simply by inserting the definition of $V^*$.
From this, it is obvious why $(0,q)$-tensors have $q$ copies of $V^*$ in the tensor product $(2)$, but $q$ copies of $V$ in the domain of the multilinear function in $(3)$.
OK, but why do you have a $V^*$ in the map in $(3)$ for each factor $V$ in the tensor product? After all, vectors are not functions, are they?
Well, in some sense they are: There is a natural linear map from $V$ to its double dual $V^{**}$, that is, the set of linear functions from $V^*$ to $K$. Indeed, for finite dimensional vector spaces, you even have that $V^{**} \cong V$. This natural map is defined by the condition that applying the image of $v$ to $\phi\in V^*$ gives the same value as applying $\phi$ to $v$. I suspect that the lecture assumes finite dimensional vector spaces. In that case, you can identify $V$ with $V^{**}$, and therefore you get
$$T^1_0V = V = V^{**} = \{T:V^*\to K\mid T \text{ is linear}\}$$
Here the second equality is exactly that identification.
Now again it should be obvious why $p$ copies of $V$ in the tensor product $(2)$ give $p$ factors of $V^*$ for the domain of the multilinear functions in $(3)$.
Edit: On request in the comments, something about the relations of those terms to the Kronecker product.
The tensor product $\color{darkorange}{\otimes}$ in $(2)$ is a tensor product not of (co)vectors, but of (co)vector spaces. The result of that tensor product describes not one tensor, but the set of all tensors of a given type. The tensors are then elements of the corresponding set. And given a basis of $V$, the tensors can then be specified by giving their coefficients in that basis.
This is completely analogous to the vector space itself. We have the vector space, $V$, this vector space contains vectors $v\in V$, and given a basis $\{e_i\}$ of $V$, we can write the vector in components, $v = \sum_i v^i e_i$.
Similarly for $V^*$, we can write each member $\phi\in V^*$ in the dual basis $\omega^i$ (defined by $\omega^i(e_j)=\delta^i_j$) as $\sum_i \phi_i \omega^i$. An alternative way to get the components $\phi_i$ is to notice that $\phi(e_k) = \sum_i \phi_i \omega^i(e_k) = \sum_i \phi_i \delta^i_k = \phi_k$. That is, the components of the covector are just the function values at the basis vectors.
This way one also sees immediately that $\phi(v) = \sum_i \phi(v^i e_i) = \sum_i v^i\phi(e_i) = \sum_i v^i \phi_i$, which is sort of like an inner product, but not exactly, because it behaves differently at change of basis.
Now let's look at a $(0,2)$ tensor, that is, a bilinear function $f:V\times V\to K$. Note that $f\in V^*\color{darkorange}{\otimes} V^*$, as $V^*\color{darkorange}{\otimes} V^*$ is by definition the set of all such functions (see eq. $(3)$). Now by being a bilinear function, one again only needs to know the values at the basis vectors, as
$$f(v,w) = f(\sum_i v^i e_i, \sum_j w^j e_j) = \sum_{i,j}v^i w^j f(e_i,e_j)$$
and therefore we can define as components $f_{ij} = f(e_i,e_j)$ and get $f(v,w)=\sum_{i,j}f_{ij}v^i w^j$.
This goes also for general tensors: A single tensor $T\in T^p_qV$ is a multilinear function $T:(V^*)^p\times V^q\to K$, and it is completely determined by the values you get when inserting basis vectors and basis covectors everywhere, giving the components
$$T^{i\ldots j}_{k\ldots l}=T(\underbrace{\omega^i,\ldots,\omega^j}_{p},\underbrace{e_k,\ldots,e_l}_{q})$$
OK, we now have components, but we have still not defined the tensor product of tensors. But that is actually quite easy:
Be $x\in T^p_qV$, and $y\in T^r_sV$. That is, $x$ is a function that takes $p$ covectors and $q$ vectors, and gives a scalar, while $y$ takes $r$ covectors and $s$ vectors to a scalar. Then the tensor product $x\color{blue}{\otimes} y$ is a function that takes $p+r$ covectors and $q+s$ vectors, feeds the first $p$ covectors and the first $q$ vectors to $x$, and the remaining $r$ covectors and $s$ vectors to $y$, and them multiplies the result. That is,
$$(x\color{blue}{\otimes} y)(\underbrace{\kappa,\ldots,\lambda,\mu,\ldots,\nu}_{p+r},\underbrace{u,\ldots,v,w,\ldots,x}_{q+s}) = x(\underbrace{\kappa,\ldots,\lambda}_p,\underbrace{u,\ldots,v}_q)\cdot y(\underbrace{\mu,\ldots,\nu}_{r},\underbrace{w,\ldots,x}_{s})$$
It is not hard to check that this function is indeed also multilinear, and therefore $x\color{blue}{\otimes} y\in T^{p+r}_{q+s}V$.
And now finally, we get to the question what the components of $x\color{blue}{\otimes} y$ are. Well, the components of $x\color{blue}{\otimes} y$ are just the function values when inserting basis vectors and basis covectors, and when you do that and use the definition of the tensor product, you find that indeed, the components of the tensor product are the Kronecker product of the components of the factors.
Also, it can be shown that $T^p_q V$ is a vector space in its own right, and therefore the $(p,q)$-tensors can be written as the linear combination of a basis that is $1$ exactly for one combination of basis vectors and basis covectors and $0$ for all other combinations. However it can then easily be seen that this is just the tensor product of the corresponding dual covectors/vectors. Since furthermore in that basis, the coefficients on the basis vectors are just the components of the tensor as introduced before, we finally arrive at the formula
$$T = \sum T^{i\ldots j}_{k\ldots l}\underbrace{e_i\color{blue}{\otimes}\dots\color{blue}{\otimes} e_j}_{p}\color{blue}{\otimes}\underbrace{\omega^k\color{blue}{\otimes}\dots,\color{blue}{\otimes}\;\omega^l}_{q}$$
I assume that $e_1,\dots,e_n$ is a basis of $V$ and that $e^1,\dots,e^n$ the associated dual basis of $V^*$.
First, let's consider the case of arbitrary (not necessarily symmetric) tensors. We note that, by linearity,
$$
T(v^{(1)}, \dots, v^{(k)}) =
T\left( \sum_{i=1}^n v^{(1)}_i e_i, \dots, \sum_{i=1}^n v^{(k)}_i e_i \right) =
T\left( \sum_{i_1=1}^n v^{(1)}_{i_1} e_i, \dots, \sum_{i_k=1}^n v^{(k)}_{i_k} e_{i_k} \right) = \\
\sum_{i_1=1}^n \cdots \sum_{i_k=1}^n v^{(1)}_{i_1} \cdots v^{(k)}_{i_k} T\left(e_{i_1}, \dots, e_{i_k} \right)
$$
Now, define the tensor $\tilde T$ by
$$
\tilde T = \sum_{i_1=1}^n \cdots \sum_{i_k=1}^n T\left(e_{i_1}, \dots, e_{i_k} \right) e^{i_1} \otimes \cdots \otimes e^{i_k}
$$
Prove that $\tilde T(v^{(1)},\dots,v^{(k)}) = T(v^{(1)},\dots,v^{(k)})$ for any $v^{(1)},\dots,v^{(k)}$. That is, $\tilde T = T$. We've thus shown that any (not necessarily symmetric) $k$-tensor can be written as a linear combination of $e^{i_1} \otimes \cdots \otimes e^{i_k}$.
The same applies for symmetric tensors. However, if $T$ is symmetric, then
$$
T\left(e_{i_1}, \dots, e_{i_k} \right) =
T\left(e_{\sigma(i_1)}, \dots, e_{\sigma(i_k)} \right)
$$
for any permutation $\sigma$. Thus, we may regroup the above sum as
$$
T = \tilde T = \sum_{i_1=1}^n \cdots \sum_{i_k=1}^n T\left(e_{i_1}, \dots, e_{i_k} \right) e^{i_1} \otimes \cdots \otimes e^{i_k} =
\\
\sum_{1 \leq i_1 \leq \cdots \leq i_k \leq n} \;
\frac 1{\alpha(i_1,\dots,i_k)}\sum_{\sigma \in S_k} T\left(e_{\sigma(i_1)}, \dots, e_{\sigma(i_k)} \right)
e^{\sigma(i_1)} \otimes \cdots \otimes e^{\sigma(i_k)} =
\\
\sum_{1 \leq i_1 \leq \cdots \leq i_k \leq n} \;
\frac 1{\alpha(i_1,\dots,i_k)}\sum_{\sigma \in S_k} T\left(e_{i_1}, \dots, e_{i_k} \right)
e^{\sigma(i_1)} \otimes \cdots \otimes e^{\sigma(i_k)} =
\\
\sum_{1 \leq i_1 \leq \cdots \leq i_k \leq n}
\frac 1{\alpha(i_1,\dots,i_k)}
T\left(e_{i_1}, \dots, e_{i_k} \right)
\underbrace{\sum_{\sigma \in S_k} e^{\sigma(i_1)} \otimes \cdots \otimes e^{\sigma(i_k)}}_{\text{basis element for } Sym^k(V)}
$$
Thus, we have expressed $T$ as a linear combination of the desired basis elements.
${\alpha(i_1,\dots,i_k)}$ counts the number of times any element $(\sigma(i_1),\dots,\sigma(i_n))$ appears in the summation over $\sigma \in S_n$. As the comment below points out, we have
$$
\alpha(i_1,\dots,i_k) = m_1! \cdots m_n!
$$
where $m_j$ is the multiplicity of $j \in \{1,\dots,n\}$ in the tuple $(i_1,\dots,i_k)$.
Best Answer
Let $\mathbb F$ be a field of characteristic different from $2$. On $V\otimes V$ you have an automorphism $T$ defined by $T(v\otimes w)=w\otimes v$ on simple tensors, since $w\otimes v$ is a bilinear expression in $(v,w)$. Note that for each $x\in V\otimes V$ you have $$ x = \frac{1}{2} \bigg(x + T(x)\bigg) + \frac{1}{2} \bigg(x - T(x)\bigg), $$ where the first summand is a symmetric tensor and the second is alternating.
Hence $V\otimes V$ is the sum of the subspaces of symmetric and alternating tensors. It is direct since the only $(2,0)$-tensor that is both symmetric and alternating is the $0$ tensor (since $\operatorname{char}(\mathbb F)\neq 2$).
The same argument applies to $V^*\otimes V^*$.
Note that the statement is false in characteristic $2$ where $v\otimes v$ is both symmetric and alternating for all $v\in V$.