Let $m$ be a square free integer and $p$ an odd prime which divides $m$. I wish to factor the ideal $(p)$ over the ring of integers of $\mathbb K = \mathbb Q(\sqrt m)$.
I know of an algorithm to use ramification theory to factor the prime.
Given a prime $p$, and a number field $\mathbb K$, to factor $(p)$ over $\mathcal O_\mathbb K$, we take $f(x)$ to be the minimal polynomial of $\alpha$ where $\mathcal O_\mathbb K = \mathbb Z[\alpha]$. We then factor $f$ modulo $p$, so that $f(x) = \prod f_i(x)^{e_i} \mod p$. We then lift these irreducible factors $f_i(x)$ to $\mathbb Z[x]$ to get polynomials say $f_i'$ and then $(p) = \prod (p, f_i'(\alpha))^{e_i}$.
The case where $m \not\equiv 1 \mod 4$ is fairly straightforward. In this case $\mathcal O_\mathbb K = \mathbb Z[\sqrt{m}]$. The minimal polynomial of $\sqrt{m}$ is $x^2 - m$ which reduces to $x^2$ modulo $p$ since $p | m$. Thus the only irreducible factor is $f_1'(x) = x$, so we have $(p) = (p, \sqrt{m})^2$.
The case where $m \equiv 1 \mod 4$ is confusing me a bit though, because in this case $\alpha = \frac{1+\sqrt{m}}{2}$, which has minimal polynomial $f(x) = x^2 - x + \frac{1-m}{4}$. I can't really see a neat way of factoring this modulo $p$. From trying some specific values of $m$ and $p$ I've conjectured that $f(x) = (x + \frac{p-1}{2})^2 \mod p$, but I haven't proved this, and this gives that $(p) = (p, \frac{p+\sqrt{m}}{2})^2$, but I would have thought the factorisation would be the same regardless of what $m$ is, and this doesn't look the same as $(p, \sqrt{m})^2$ to me.
Because $p$ is odd and $p\mid m$, we have $\dfrac{1-m}4\equiv\dfrac14\pmod p$. Therefore $$ x^2-x+\frac{1-m}4\equiv x^2-x+\frac14=(x-\frac12)^2\pmod p $$ confirming your hunch.
There is nothing wrong with your other calculations either. Namely the ideals $$ \mathfrak{p}_1=(p,\sqrt m) $$ and $$ \mathfrak{p}_2=(p,\frac{p+\sqrt m}2) $$ are equal. Trivially $\sqrt m\in \mathfrak{p}_2$ so $\mathfrak{p}_1\subseteq\mathfrak{p}_2.$ On the other hand, the calculation $$ \frac{1+\sqrt m}2\cdot p-\frac{p-1}2\cdot\sqrt m=\frac{p+\sqrt m}2 $$ expresses $(p+\sqrt m)/2$ as an $\mathcal{O}_K$-linear combination of generators of $\mathfrak{p}_1$. Hence also $\mathfrak{p}_2\subseteq\mathfrak{p}_1.$