I have a vector:
$w$ = $(1, 1)$
That I need to express as a linear combination of orthonormal basis vectors:
$u$ = $(\frac{1}{\sqrt5},\frac{3}{\sqrt5})$ and $v$ = $(\frac{2}{\sqrt5},\frac{1}{\sqrt5})$, which have been normalised from $u$ = $(1, 3)$ and $v$ = $(2, 1)$.
Note: I am using the inner product of:
$(u, v)$ = $2$$v_1$$u_1$ - $v_1$$u_2$ - $v_2$$u_1$ + $v_2$$u_2$
How do I go about doing this?
You seek coefficients $a$ and $b$ such that $au + bv = w$. If we express $u$, $v$, and $w$ as column vectors, and define the matrix $A = [u | v]$ whose columns are $u$ and $v$, then $au + bv$ can be written as $$A\begin{pmatrix}a \\ b \end{pmatrix}$$ You want this equal to $w$, so $$A\begin{pmatrix}a \\ b\end{pmatrix} = w$$ If we multiply both sides by $A^{-1}$, we obtain the desired coefficients: $$\begin{pmatrix}a \\ b\end{pmatrix} = A^{-1}w$$
This is particularly easy for orthonormal $u$ and $v$, for in that case we have $A^{-1} = A^t$. However, as noted in the comments, your $u$ and $v$ are not orthogonal, hence not orthonormal, assuming the standard inner product.
Another approach is to start with $$au + bv = w$$ and take the inner product of both sides with $u$ and with $v$. Assuming $u$ and $v$ are orthonormal, taking the inner product of both sides with $u$ gives us $$a\langle u,u\rangle + b\langle v,u \rangle = \langle w,u \rangle$$ which simplifies to $$a = \langle w,u \rangle$$ since $\langle u,u\rangle = 1$ and $\langle v,u \rangle = 0$.
Similarly, taking the inner product of both sides of $au + bv = w$ with $v$ gives us $$b = \langle w, v \rangle$$
Note that the first technique works even if $u$ and $v$ are not orthonormal, but requires inverting a matrix. The second technique is computationally easier but requires orthonormality of $u$ and $v$.
Since you are using a nonstandard inner product, you will not have $A^{-1} = A^t$ in the first method. So the second method will be easier, but be sure to use your inner product and not the standard one.
Applying method 1 to your situation, we have $$A = \frac{1}{\sqrt{5}}\begin{pmatrix}1 & 2 \\ 3 & 1 \end{pmatrix}$$ so $$A^{-1} = \frac{1}{\sqrt{5}}\begin{pmatrix}-1 & 2 \\ 3 & -1\end{pmatrix}$$ and therefore $$\begin{pmatrix}a \\ b\end{pmatrix} = A^{-1}\begin{pmatrix}1 \\ 1\end{pmatrix} = \frac{1}{\sqrt{5}}\begin{pmatrix}1 \\ 2\end{pmatrix}$$ Therefore $$a = \frac{1}{\sqrt{5}}$$ and $$b = \frac{2}{\sqrt{5}}$$
Applying method 2 to your situation, we have $$\begin{aligned} a &= \langle w, u\rangle \\ &= 2u_1 w_1 - u_1 w_2 - u_2 w_1 + u_2 w_2 \\ &= \frac{1}{\sqrt{5}} (2(1 \cdot 1) - 1\cdot 1 - 3\cdot 1 + 3\cdot 1) \\ &= \frac{1}{\sqrt{5}} \end{aligned}$$ and $$\begin{aligned} b &= \langle w, v\rangle \\ &= 2v_1 w_1 - v_1 w_2 - v_2 w_1 + v_2 w_2 \\ &= \frac{1}{\sqrt{5}} (2(2 \cdot 1) - 2\cdot 1 - 1\cdot 1 + 1\cdot 1) \\ &= \frac{2}{\sqrt{5}} \end{aligned}$$