The combined state space of the two qubits is spanned by $|00\rangle,|01\rangle,|10\rangle,|11\rangle$. You can consider $|xy\rangle$ as shorthand for the formal product $|x\rangle|y\rangle$ (which is not equal to $|y\rangle|x\rangle$ if $x\ne y$), and using that you find that the combined state is $ac|00\rangle + ad|01\rangle + bc|10\rangle + bd|11\rangle$. This is the tensor product of the states, which is an element of the tensor product of the two state spaces.
Note that this is not an entangled state, by definition. It is entangled exactly when it is not of this form. A general 2-qubit state is not of this form and can not be written as the tensor product of 1-qubit states (only as a linear combination of such separable states). To find out whether it is separable, and to separate them in that case, amounts to solving a system of equations.
In general, if a 1-qubit gate maps $|0\rangle$ to $a|0\rangle + b|1\rangle$, then, if it is applied to (e.g.) the second qubit of a pair, then $|00\rangle$ is mapped to $a|00\rangle + b|01\rangle$, etc. This is the tensor product of the identity and the 1-qubit gate. For the corresponding matrices this is the Kronecker product.