To see that a linear map of vector spaces is a $(1,1)$-tensor, realize that such an object eats a vector $X$ and covector $\omega$ (linear form to $\mathbb R$!) and gives you a number, i.e. $$\mathbf T(X,\omega)=\sum_{i,j} X^i\omega_j\mathbf T(\partial_i,dx^j)=\sum_{i,j} \omega_jT^j_iX^i,\;\text{where }T^j_i:=\mathbf T(\partial_i,dx^j)\in\mathbb R.$$
In the previous formula one can see that a $(1,1)$-tensor "is" a matrix $T^j_i$ which takes a vector $X$ and gives a vector $\sum_i T^j_iX^i$, which is precisely the matrix characterization of a linear map between vector spaces, precisely the linear map $\mathbf T(\_,\cdot):V\rightarrow V$ given by $X\mapsto\mathbf T(X,\cdot)$.
Thinking of tensors as multi-dimensional arrays is indeed a conceptual understanding of what they are as long as you imagine them acting linearly on vectors. You may be interested in this long answer I gave at this other question concerning the concept and construction of covector in manifolds (and thus generalizing to tensors). You can think that your multi-dimensional multi-linear arrays have their entries smoothly dependent on the points of the manifold, in such a way that the whole object remains linear when acting on smooth vector fields (i.e. sections of the tangent bundle). For this to be true, their components have to transform in a particular way (generalizing the transformation law derived at the linked answer above).
If covectors are smoothly varying linear forms $\omega\vert_p :T_pM\rightarrow\mathbb R$ such that $\omega (aX+bY)=a\omega(X)+b\omega(Y)\in\mathbb R$ for all $a,b\in\mathbb R$ and any smooth vector fields $X,Y\in TM$, then they are completely determined, by linearity (check!), by their action on any coordinate basis of any chart: $$\omega(\partial_i)=:\omega_i\Rightarrow \omega (X)=\sum_i X^i\omega(\partial_i)=\sum_i X^i\omega_i\,.$$
Since $X^i$ and $\omega_i$ are by definition of vectors and covectors smooth scalar fields on $M$, we have checked that indeed such $\omega$ are the linear forms $TM\rightarrow\mathbb R$. Now, define covariant $k$-tensors to be similarly generalized from punctual multi-linear forms $\Omega\vert_p:\otimes^k T_pM\rightarrow\mathbb R$, that is to say, linear functionals on $k$ vectors fields: $$\Omega(aX_1+bY_1,X_2,...,X_k) =a\Omega(X_1,...,X_k)+ b\Omega(X_1,...,X_k)\text{ and similarly for the other slots}.$$
Because of this multi-linearity, their definition as action on a number of vectors reduces to action on coordinate basis: $$\Omega(\partial_{i_1},...,\partial_{i_k}) =\Omega_{i_1...i_k}\Rightarrow \Omega(X_1,...,X_k)=\sum_{i_1,...,i_k}X^{i_1}_1\cdots X^{i_k}_k\Omega_{i_1...i_k}\,.$$
In order to extend this algebraic (co)-tensors at every point to tensor fields on the manifold, their multi-array components $\Omega_{i_1...i_k}(P)$ must be smooth functions on the points $P\in M$, i.e. $\Omega_{i_1...i_k}:M\rightarrow\mathbb R$, but for the whole array to behave coherently and multi-linearly, since vectors transform between charts by the basis transformation, their components must patch together as: $$\partial'_i=\sum_j\frac{\partial x^j}{\partial y_i}\partial_j\Rightarrow \Omega'_{i_1...i_k}:=\Omega(\partial'_{i_1},...,\partial'_{i_k})=\sum_{j_1,...,j_k}\frac{\partial x^{j_1}}{\partial y_{i_1}}\cdots\frac{\partial x^{j_k}}{\partial y_{i_k}}\Omega_{j_1...j_k}\,.$$
This is the reason to the, too often, confusing fact that the components of a covector transform as the basis of vectors, and the components of a vector as the basis of covectors!
If you generalize this to include contravariant $k$-tensors $A\vert_p:\otimes^k T_p^*M\rightarrow\mathbb R$, it is easy to deduce that their transformation between charts has the oposite jacobian matrices. Finally you put together all this to define $(r,s)$-tensors $T\vert_p:\otimes^r T_pM\otimes^s T_p^*M\rightarrow\mathbb R$, which are multi-linear objects which take $r$ vectors and $s$ covectors and give numbers, make them into tensor fields by letting their array components to vary smoothly on $M$ and ensuring their patching on overlaping charts to preserve linearity by:
$$T'\,^{i_1...i_s}_{j_1...j_r}=\sum_{l_1,...,l_s}\sum_{k_1,...,k_r}\frac{\partial x^{k_1}}{\partial y_{j_1}}\cdots\frac{\partial x^{k_r}}{\partial y_{i_k}}\cdot\frac{\partial y^{i_1}}{\partial x_{l_1}}\cdots\frac{\partial y^{i_r}}{\partial x_{l_k}}T^{l_1...l_s}_{k_1...k_r}\,.$$
Therefore, indeed you can think of tensors as general multi-linear multi-dimensional arrays of smooth real functions on every chart of your manifold, such that all of them patch together nicely on the charts' intersections (think of charts as coordinate systems, like observers in physics, so any of them has a bunch of functions making these arrays, and any two observers agree they are talking about the same array-object by checking that their action on any input is the same regardless of their coordinate transformations). So actually, you end up realizing that you could have defined tensors as sections of the tensor product bundle of several copies of the tangent and cotangent bundles, since that is the most geometric, intrinsic and coordinate-independent definition possible. By taking only anti-symmetric covariant tensors you get the differential forms of the other answer.
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
$$T = \sum T^{i\ldots j}_{k\ldots l}\underbrace{e_i\color{blue}{\otimes}\dots\color{blue}{\otimes} e_j}_{p}\color{blue}{\otimes}\underbrace{\omega^k\color{blue}{\otimes}\dots,\color{blue}{\otimes}\;\omega^l}_{q}$$
Best Answer
Note that even Wikipedia's definition is not the most general. The most general definition (that I know of) for a (real) tensor is a multilinear map $$ T: V_1 \times V_2 \times \cdots \times V_k \rightarrow \mathbb{R},$$ where $V_1, \dots, V_k$ are (not necessarily the same) vector spaces. And such generalized tensors definitely do appear in differential geometry and topology.
I also want to note that these simple definitions are also very misleading in the sense that the tensors that arise in geometry and topology all have symmetries or anti-symmetries in the indices. This leads to most of the headaches when learning how to deal with tensors. In fact, if you get serious with tensors, the notation can become quite a challenge. Penrose developed a particularly creative approach to dealing with this.
However, when you're first learning differential geometry and topology, almost all of the tensors encountered are indeed $(0,q)$ tensors. That's because either you're working with only exterior differential forms or you're working with a Riemannian manifold, where the metric can be used to convert any $(p,q)$-tensor into $(0,p+q)$-tensor.
It's good to know the definitions of general tensors (as described above) and $(p,q)$ tensors, because they are sometimes necessary or useful. But it's probably best to avoid devoting much attention to them in introductory courses and textbooks.
ADDED: An extremely useful canonical isomorphism is $$ \operatorname{Hom}(V,W) = V^*\otimes W $$ where $V$ and $W$ are vector spaces and $\operatorname{Hom}(V,W)$ is the space of linear maps from $V$ to $W$. Used with care, this means that you can view a $(p,q)$ tensor $T \in V_1^*\otimes\cdots\otimes V_q^*\otimes V_{q+1} \otimes \cdots \otimes V_{p+q}$ as a multilinear map $$ T: V_{1}\times\cdots\times V_{q} \rightarrow V_{q+1}\otimes\cdots\otimes V_{p+q}. $$ In short, to avoid working with $(p,q)$-tensors directly, it can be useful to view them as maps from $(q,0)$-tensors to $(p,0)$-tensors instead. Or, equivalently, as $(V_{q+1}\otimes\cdots\otimes V_{p+q})$-valued $(0,q)$-tensors.