Why is quantum entanglement so important in the context of quantum computation?
Both of these terms depend on what basis you're in, which can make them a little arbitrary. For example, I think of a single particle in the state $\left|\uparrow\right> + \left|\downarrow\right>$ as in a superposition, though someone else who prefers the $\hat x$ basis may disagree and call it their eigenstate (we're both right). Similarly, entanglement is something we only understand if we work in the single-particle basis. There, we define an entangled state as a state that cannot be written as a product of single-particle states. By a single-particle state, I mean something that can be written as $\left|\psi\right>_1 \otimes \left|\psi \right>_2$ for two particles.
For example, $\left|\uparrow\right>_1 \left|\uparrow\right>_2$ is not entangled, since it's just the product of $\left|\uparrow\right>_1$ and $\left|\uparrow\right>_2$. This is a very correlated state, but it's not entangled. Here, I'm using the subscript to denote which particle. An example of a state that isn't entangled is:
$$ \left|\uparrow\right>_1 \left|\uparrow\right>_2 + \left|\uparrow\right>_1 \left|\downarrow\right>_2 + \left|\downarrow\right>_1 \left|\uparrow\right>_2 + \left|\downarrow\right>_1 \left|\downarrow\right>_2$$
this is because I can write it as:
$$(\left|\uparrow\right>_1 + \left|\downarrow\right>_1) \otimes (\left|\uparrow\right>_2 + \left|\downarrow\right>_2)$$
On the other hand, the following states are entangled:
$$ \left|\uparrow\right>_1 \left|\uparrow\right>_2 + \left|\uparrow\right>_1 \left|\downarrow\right>_2 + \left|\downarrow\right>_1 \left|\uparrow\right>_2 - \left|\downarrow\right>_1 \left|\downarrow\right>_2$$ $$ \left|\uparrow\right>_1 \left|\uparrow\right>_2 + \left|\downarrow\right>_1 \left|\downarrow\right>_2$$
The latter is known as a Bell state. You cannot write either of these states as a generic $\left|\psi\right>_1 \otimes \left|\psi \right>_2$: go ahead and try!
A more practical metric I use is: does the result of measuring the state of one particle change my expectation of the state of another? If yes, the particles must be entangled. (Warning: if the answer is no, it doesn't rule out entanglement! You might need a more clever experiment). For example, if I start with $\left|\uparrow\right>_1 \left|\downarrow\right>_2$, and I know I will measure that the second particle is $\downarrow$, regardless of what basis I measure the first particle in. However, if I start with the state $\left|\uparrow\right>_1 \left|\uparrow\right>_2 + \left|\downarrow\right>_1 \left|\downarrow\right>_2$, I have $50-50$ odds measuring that the second particle is $\uparrow$ or $\downarrow$. If I measure that the first particle is $\uparrow$, suddenly I know that I'll measure the second particle is $\uparrow$. I've learned something about the second particle only by measuring the first. A measurement on the first particle has changed the odds on the outcome of measuring the second, the hallmark of entanglement.
Side note: If a friend places two marbles in a bag, and promises that both are red or both are blue, you'd get a similar result as my second experiment. But this is not entanglement, that's just a classical lack of knowledge! My proposed experiment to measure entanglement of $\left|\uparrow\right>_1 \left|\uparrow\right>_2 + \left|\downarrow\right>_1 \left|\downarrow\right>_2$, by measuring each particle in the $\hat z$ basis is convincing only if you know that you started in a pure, quantum mechanical state. In reality you need a more clever experiment to show that you have an entangled state. Examples include Bell's inequality and the CHSH inequality.
An entangled state is essentially a superposition of states of different qubits, in such a way that they cannot be factored into a tensor product of individual states, such as the Bell state $$|\psi\rangle=\frac{1}{\sqrt{2}}\left(|00\rangle+|11\rangle\right).$$ In such a state both qubits are correlated in the specific sense that they exhibit non-classical correlations that can be used to break Bell inequalities inaccessible to states with correlations achievable under local operations and classical communication.
It is important to use entanglement - rather than superposition - as a resource in quantum computing, because if you only allow single-qubit superposition states then the computation can be efficiently simulated by a classical computer. More specifically, since you can write all single-qubit states as $$|\psi\rangle=\cos(\theta)|0\rangle+e^{i\phi}\sin(\theta)|1\rangle,$$ with two real parameters, you can change each qubit for $2n$ classical bits, to simulate the quantum computation to $n$-bit accuracy. This is a polynomial overhead and thus turns efficient computations into efficient computations.
It's important to note that entanglement need not be all there is to quantum computing as some non-entangled states can exhibit non-classical behaviour. This is one reason why quantum discord is being championed as a better figure of merit for the "nonclassicalness" of a state by some.
Generally superposition refers to a single particle having a combination ('superposition') of two states. For example, a photon having a combination of vertical and horizontal polarization.
Entanglement generally refers to different particles having correlated quantum states. For example, the spin of two (physically separated) electrons being antiparallel.
A qubit can exhibit both superposition and entanglement, which classical 'bits' do not.
Entanglement can be used as a tool in quantum computing, for example in 'superdense' coding---which is able to transport two bits of classical information via a single entangled qubit.