On the number of quadratic residues $\pmod{pq}$ where$p$ and $q$ are odd primes. I have read that the formula for the number of quadratic residues $\pmod{pq}$ for odd primes $p$ and $q$ is $\frac{(p-1)(q-1)}{4}$. Is this the case, and if it is, why is it the case and how would one going about proving it? More generally, is there a formula for the number of quadratic residues for any composite modulus $m$ (and if so, why is it so and how would I prove that the formula holds)?

For any $x$ between $1$ and $pq-1$ which is relatively prime to $pq$, let $f(x)$ be the remainder when $x^2$ is divided by $pq$. We show that the function $f$ is $4$-to-$1$. That  yields your formula. 
To show that $f$ is $4$-to-$1$, note that $x^2\equiv y^2 \pmod{pq}$ if and only if $(zx)^2\equiv 1\pmod{pq}$, where $z$ is the multiplicative inverse of $y$. 
Now all we need to do is to show that the congruence $w^2\equiv 1\pmod{pq}$ has precisely $4$ solutions. Note that $w$ is a solution of the congruence if and only if $w^2\equiv 1\pmod{p}$ and $w^2\equiv 1\pmod{q}$. Each of these congruences has $2$ solutions, so using the Chinese Remainder Theorem we find that the congruence $w^2\equiv 1\pmod{q}$ has $4$ solutions. 
Essentially the same argument shows that if $n=p_1^{a_1}\cdots p_k^{a_k}$, where the $p_i$ are distinct odd primes, there are $\frac{\varphi(n)}{2^k}$ numbers $x$ in the interval $[1,n-1]$, relatively prime to $n$, which are squares modulo $n$. Here $\varphi$ is the Euler $\varphi$-function.
Remark: Even $n$ are a little more complicated, one has to treat separately the cases where the highest power of $2$ that divides $n$ is $2^1$ or $2^2$ and the cases where the highest power is $2^k$ for some $k\ge 3$. 
Added: In case the reduction to the congruence $w^2\equiv 1 \pmod{pq}$ seems mysterious, here is a more elementary approach. Let $a$ be fixed. We want to find how many solutions there are to the congruence $x^2\equiv a^2\pmod{pq}$. 
Rewrite this as $(x-a)(x+a)\equiv 0 \pmod{pq}$. The congruence holds if one of $x-a$ or $x+a$ is divisible by $p$, and one of $x-a$ or $x+a$ is divisible by $q$. The $4$ solutions are obtained by setting (1) $x\equiv a \pmod{p}$, $x\equiv a \pmod{q}$; (2) $x\equiv -a \pmod{p}$, $x\equiv -a \pmod{q}$; (3) $x\equiv a \pmod{p}$, $x\equiv -a \pmod{q}$; (4) $x\equiv -a \pmod{p}$, $x\equiv a \pmod{q}$. Each of these systems of $2$ congruences has a unique solution modulo $pq$ by the Chinese Remainder Theorem.  
