I think using examples would convey my question the fastest.
Let there be a set $A = \{ 1,2,3,5,10 \}$. Let $R$ be relation such that $a | b$.
Note that in this case $a | b$ is what I mean by "condition" which I mentioned in the question
Thus $R= \{ (1,2), (1,3), (1,5), (1,10), (2,10), (5,10) \}$
One possible total order $T$ would be $T= \{ (1,2), (1,10) \}$ since $1,2,10$ are comparable,ie ($1$⪯$2$⪯$10$) and is partial order.
Both $T$ and $R$ are related by the same condition, $a | b$... right?
But then my school showed that another possible total order $T$ to be $1 ≺ 2 ≺ 3 ≺ 5 ≺ 10$ ie $T= \{ (1,2), (1,3),...(2,3),(2,5)... (2,10), (5,10) \}$
But how is $2|3$ or $2|5$? Is the condition no longer $a | b$?
My question essentially boils down to the title. So can the partial and total order not have the same "condition" at all?
You're right: the condition (that is to say, the relation itself) is different in the second case: it's simply the usual $\le$.
Nothing prevents us to consider two different relations on the same set. Nevertheless the chain $1,2,3,5,10$ is not a chain with respect to the first relation.
Note that though these are two different relations, they have a connection, namely the first one is included in the second one, meaning that $a|b\implies a\le b$ (at least for positive integers).