Hadamard gate - resulting state For a single qubit Hadamard gate, the representing matrix is $$\frac{1}{\sqrt{2}}\begin{bmatrix}1&1\\1&-1\end{bmatrix}$$
And if you apply this to the state $|0\rangle$ you get $$\frac{|0\rangle+|1\rangle}{\sqrt{2}} \, .$$
This state can be rewritten as
$$
\frac{\begin{bmatrix}0\\1\end{bmatrix}+\begin{bmatrix}1\\0\end{bmatrix}}{\sqrt{2}}
\quad \text{or} \quad
\frac{\begin{bmatrix}1\\1\end{bmatrix}}{\sqrt{2}}
\quad \text{or} \quad
\frac{1}{\sqrt{2}} \begin{bmatrix}1\\1\end{bmatrix}
\quad \text{or} \quad \begin{bmatrix}\frac{1}{\sqrt2}\\\frac{1}{\sqrt2}\end{bmatrix}
$$
right? 
If so, using this, let's say you have the superposition $\alpha |0\rangle + \beta |1\rangle$. Can you then plug it in and get $\begin{bmatrix}\frac{\beta}{\sqrt2}\\\frac{\alpha}{\sqrt2}\end{bmatrix}$? So if you started with a qubit represented by the vector $\begin{bmatrix}0.5\\0.75\end{bmatrix} \, ,$
say, then you'd end up with
$\begin{bmatrix}\frac{0.5}{\sqrt2}\\\frac{0.75}{\sqrt2}\end{bmatrix} \, ,$ right?
Edit: So, overall, I'd say I'm having trouble figuring out exactly why you'd write it as $\frac{|0\rangle+|1\rangle}{\sqrt{2}}$ when you could write it as $\begin{bmatrix}\frac{\beta}{\sqrt2}\\\frac{\alpha}{\sqrt2}\end{bmatrix}$, which strikes me as easier to do calculations with (though again, that's just me).
 A: The hadamard gate: 

$\hat{H} = \frac{1}{\sqrt{2}} \begin{bmatrix}1&1\\1&-1\end{bmatrix}$ 

It's (maybe) more easily understood as: 

$\hat{H} = \frac{1}{\sqrt{2}} (\lvert 0 \rangle \langle 0 \lvert + \lvert 0 \rangle \langle 1 \lvert + \lvert 1 \rangle \langle 0 \lvert - \lvert 1 \rangle \langle 1 \lvert)$

So it's straight forward to obtain the result: 

$\hat{H} \lvert 1 \rangle = \frac{1}{\sqrt{2}} (\lvert 0 \rangle \langle 0 \lvert + \lvert 0 \rangle \langle 1 \lvert + \lvert 1 \rangle \langle 0 \lvert - \lvert 1 \rangle \langle 1 \lvert) \lvert 1 \rangle = \frac{1}{\sqrt{2}} (\lvert 0 \rangle - \lvert 1 \rangle$)

Where we used that $\langle 0 \lvert 1 \rangle = 0$. In the same fashion you could calculate the result of $\hat{H}(\alpha \lvert 0 \rangle + \beta \lvert 1 \rangle)$ where you obtain:

$\hat{H}(\alpha \lvert 0 \rangle + \beta \lvert 1 \rangle) = \frac{1}{\sqrt{2}}(\lvert 0 \rangle (\alpha + \beta) + \lvert 1 \rangle (\alpha - \beta))$

If you are willing to use the representation of the kets, your state will be: 

$\alpha \lvert 0 \rangle + \beta \lvert 1 \rangle = \alpha\begin{bmatrix}1\\0\end{bmatrix} + \beta \begin{bmatrix}0\\1\end{bmatrix} = \begin{bmatrix}\alpha \\ \beta \end{bmatrix}$ 

And then it's only a matrix multiplication
