Suppose I have a simple pendulum and I want to calculate its acceleration when the bob reaches the maximal angle. I usually choose my axes such that the y-axis will be parallel to the rope. Then the acceleration would be $g \sin\alpha$ (Because: $mg\sin\alpha=ma$). However, if I choose the y-axis to be parallel to the gravitational force ($mg$), the acceleration equals to $g\tan\alpha$ (Because: (1): $T\cos\alpha=mg$ and (2): $T\sin\alpha=ma$ where $T$ is the tension force (centripetal force) and $\alpha$ is the given angle). Obviously, these two accelerations aren't the same. So my question - which one is right? And why the other one is wrong?
The total gravitational force is $mg$ and this force is the hypotenuse of a triangle whose sides are $mg\cos\alpha$ in the direction of the rope and $mg\sin\alpha$ in the orthogonal direction. So the latter, $mg\sin\alpha$, is the force (mass times acceleration) along the circular orbit because the centripetal part of the force is cancelled by the string's tension.
My guess is that instead of dividing the gravitational force into the two orthogonal components, you are trying to divide the tension force into two components and identify one of these two components with the gravitational force.
The plan wants to divide the tension $T$ into two directions – I suppose it's the horizontal and vertical direction. In the vertical direction, you think to have the cancellation between $mg$ and $T\cos\alpha$ and in the horizontal direction, between $T\sin\alpha$ and $ma$.
However, you forgot that in the vertical direction, there's a contribution to the force from the acceleration, too. So the right conditions in this coordinate system are actually $$T\cos\alpha = mg-ma\sin\alpha, \quad T\sin\alpha = ma\cos\alpha$$ Note that instead of $ma$ from your formulae, I had to write $ma\sin\alpha$ and $ma\cos\alpha$ because the acceleration is a vector in a direction that is neither vertical nor horizontal. Multiply the first (correct) equation by $\sin\alpha$, the second one by $\sin\alpha$, and subtract them in order to eliminate $T$. You will get $$ 0 = T(\sin\alpha\cos\alpha-\cos\alpha\sin\alpha) =\\ = mg\sin\alpha - ma\sin^2\alpha-ma\cos^2\alpha=mg\sin\alpha-ma $$ which gives you $a=g\sin\alpha$ again. In other words, you have forgotten to realize that the acceleration $a$ is in a direction tilted by $\alpha$ so if you rewrite the vectors in the horizontal and vertical components, you have to rotate it into $a\sin\alpha$ and $a\cos\alpha$ with the right signs.