For a simple example suppose you have two qubits in definite states $|0\rangle$ and $|0\rangle$. The combined state of the system is $|0\rangle\otimes |0\rangle$ or $|00\rangle$ in shorthand.
Then if we apply the following operators to the qubits (image is cut from superdense coding wiki page), the resulting state is an entangled state, one of the bell states.
First in the image we have the hadamard gate acting on the first qubit, which in a longer form is $H\otimes I$ so that it is the identity operator on the second qubit.
The hadamard matrix looks like
$$H=\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$$ where the basis is ordered $\{|0\rangle,|1\rangle\}$.
So after the hadamard operator acts the state is now
$$(H\otimes I)(|0\rangle\otimes|0\rangle)=H|0\rangle\otimes I |0\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)\otimes (|0\rangle)=\frac{1}{\sqrt{2}}(|00\rangle+|10\rangle)$$
The next part of the circuit is a controlled not gate, which only acts on the second qubit if the first qubit is a $1$.
You can represent $CNOT$ as $|0\rangle\langle0|\otimes I+|1\rangle\langle1|\otimes X$, where $|0\rangle\langle0|$ is a projection operator onto the bit $0$, or in matrix form $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$. Similarly $|1\rangle\langle1|$ is $\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$.
The $X$ operator is the bit flip operator represented as $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$.
Overall the $CNOT$ matrix is $\begin{pmatrix} 1 & 0 &0 & 0 \\ 0 & 1 &0 & 0 \\ 0 & 0 &0 & 1 \\0 & 0 &1 & 0 \\\end{pmatrix}$
When we apply the $CNOT$ we can either use matrix multiplication by writing our state as a vector $\begin{pmatrix}\frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \\0 \end{pmatrix}$, or we can just use the tensor product form.
$$CNOT (\frac{1}{\sqrt{2}}(|00\rangle+|10\rangle))=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$$
We see that for the first part of the state $|00\rangle$ the first bit is $0$, so the second bit is left alone; the second part of the state $|10\rangle$ the first bit is $1$, so the second bit is flipped from $0$ to $1$.
Our final state is $$\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$$ which is one of the four Bell states which are maximally entangled states.
To see what it means for them to be entangled, notice that if you were to measure the state of the first qubit say, if you found out that it was a $0$ it immediately tells you the second qubit also has to be a $0$, because thats our only possibility.
Compare to this state for instance:
$$\frac{1}{2}(|00\rangle+|01\rangle+|10\rangle+|11\rangle).$$
If you measure that the first qubit is a zero, then the state collapses to $\frac{1}{\sqrt{2}}(|00\rangle+|01\rangle)$, where there is still a 50-50 chance the second qubit is a $0$ or a $1$.
Hopefully this gives an idea how states can be entangled. If you want to know a particular example, like entangling photons or electrons etc, then you would have to look into how certain gates can be implemented, but still you might write the mathematics the same way, the $0$ and $1$ might represent different things in different physical situations.
Update 1: Mini Guide to QM/QC/Dirac notation
Usually there's a standard computational (ortho-normal) basis for a single qubit which is $\{|0\rangle,|1\rangle\}$, say $\mathcal{H}=\operatorname{span}\{|0\rangle,|1\rangle\}$ is the vector space.
In this ordering of the basis we can identify $|0\rangle$ with $\begin{pmatrix} 1\\ 0 \end{pmatrix}$ and $|1\rangle$ with $\begin{pmatrix} 0\\ 1 \end{pmatrix}$. Any single qubit operator then can be written in matrix form using this basis. E.g. a bit flip operator $X$ (after pauli-$\sigma_x$) which should take $|0\rangle\mapsto |1\rangle$ and $|1\rangle \mapsto |0\rangle$, can be written as $\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$, the first column of the matrix is the image of the first basis vector and so on.
When you have multiple say $n$-qubits they should belong to the space $\mathcal{H}^{\otimes n}:=\overbrace{\mathcal{H}\otimes\mathcal{H}\otimes\cdots\otimes \mathcal{H}}^{n-times}$. A basis for this space is labelled by strings of zeros and ones, e.g. $|0\rangle\otimes|1\rangle\otimes |1\rangle\otimes\ldots \otimes|0\rangle$, which is usually abbreviated for simplicity as $|011\ldots0\rangle$.
A simple example for two qubits, the basis for $\mathcal{H}^{\otimes 2}=\mathcal{H}\otimes \mathcal{H}$, is $\{|0\rangle\otimes|0\rangle,|0\rangle\otimes|1\rangle,|1\rangle\otimes|0\rangle,|1\rangle\otimes|1\rangle\}$ or in the shorthand $\{|00\rangle,|01\rangle,|10\rangle,|11\rangle\}$.
There's different ways to order this basis in order to use matrices, but one natural one is to order the strings as if they are numbers in binary so as above. For example for $3$ qubits you could order the basis as $$\{|000\rangle,|001\rangle,|010\rangle,|011\rangle,|100\rangle,|101\rangle,|110\rangle,|111\rangle\}.$$
The reason why this can be useful is that it corresponds with the Kronecker product for the matrices of the operators. For instance, first looking at the basis vectors:
$$|0\rangle\otimes |0\rangle=\begin{pmatrix} 1\\ 0 \end{pmatrix}\otimes \begin{pmatrix} 1\\ 0 \end{pmatrix}:=\begin{pmatrix} 1\cdot\begin{pmatrix} 1\\ 0 \end{pmatrix} \\ 0\cdot\begin{pmatrix} 1\\ 0 \end{pmatrix} \end{pmatrix}=\begin{pmatrix} 1\\ 0\\0\\0 \end{pmatrix}$$
and
$$|0\rangle\otimes |1\rangle=\begin{pmatrix} 1\\ 0 \end{pmatrix}\otimes \begin{pmatrix} 0\\ 1 \end{pmatrix}:=\begin{pmatrix} 1\cdot\begin{pmatrix} 0\\ 1 \end{pmatrix} \\ 0\cdot\begin{pmatrix} 1\\ 0 \end{pmatrix} \end{pmatrix}=\begin{pmatrix} 0\\ 1\\0\\0 \end{pmatrix}$$
and similarly
$$|1\rangle\otimes |0\rangle=\begin{pmatrix} 0\\ 0\\1\\0 \end{pmatrix},\quad |1\rangle\otimes |1\rangle=\begin{pmatrix} 0\\ 0\\0\\1 \end{pmatrix}$$
If you have an operator e.g. $X_1X_2:=X\otimes X$ which acts on two qubits and we order the basis as above we can take the kronecker product of the matrices to find the matrix in this basis:
$$X_1X_2=X\otimes X=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\otimes \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0\cdot\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} & 1\cdot\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\\ 1\cdot\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} & 0\cdot \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \end{pmatrix}=\begin{pmatrix} 0 &0&0&1\\ 0 &0&1&0\\0 &1&0&0\\1 &0&0&0\\ \end{pmatrix}$$
If we look at the example of $CNOT$ above given as $|0\rangle\langle0|\otimes I+|1\rangle\langle1|\otimes X$.$^*$ This can be computed in matrix form as $\begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}\otimes \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}+\begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix}\otimes\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$, which you can check is the $CNOT$ matrix above.
It's worthwhile getting used to using the shorthands and the tensor products rather than converting everything to matrix representation since the computational space grows as $2^n$ for $n$-qubits, which means for three cubits you have $8\times 8$ matrices, $4$-qubits you have $16\times 16$ matrices and it quickly becomes less than practical to convert to matrix form.
Aside$^*$: There are a few common ways to use dirac notation, to represent vectors like $|0\rangle$; dual vectors e.g. $\langle 0|$, inner product $\langle 0|1\rangle$ between the vectors $|0\rangle$ and $|1\rangle$; operators on the space like $X=|0\rangle\langle1|+|1\rangle\langle0|$.
An operator like $P_0=|0\rangle\langle0|$ is a projection operator is a (orthogonal) projection operator because it satisfies $P^2=P$ and $P^\dagger=P$.
