Probability of drawing the Jack of Hearts?
The Hard Way
The probability that the first card is not the Jack of Hearts is $\frac {51}{52}$ so the probability that the first card is not the Jack of Hearts and the second card is the Jack of Hearts is $\frac {51}{52}\times \frac 1{51}$.
The probability that the first card is the Jack of Hearts is $\frac 1{52}$ so the probability that the first card is the Jack of Hearts and the second card is the Jack of Hearts is $\frac 1{52}\times 0$.
So the total probability that the second card is Jack of Hearts is:
The probability that the second card is after the first card is not +
The probability that the second card is after the first card already was
$$= \frac {51}{52}\times \frac 1{51} + \frac 1{52}\times 0 = \frac 1{52} + 0 = \frac 1{52}$$
That was the hard way.
The Easy Way
The probability that any specific card is any specific value is $\frac 1{52}$. It doesn't matter if it is the first card, the last card, or the 13th card. So the probability that the second card is the Jack of Hearts is $\frac 1{52}$. Picking the first card and not looking at, just going directly to the second card, putting the second card in an envelope and setting the rest of the cards on fire, won't make any difference; all that matters is the second card has a one in $52$ chance of being the Jack of Hearts.
Any thing else just wouldn't make any sense.
The thing is throwing in red herrings like "what about the first card?" doesn't change things and if you actually do try to take everything into account, the result, albeit complicated, will come out to be the same.
I'm confused by this question because if the card you removed from the pile was the Jack of Hearts then the probability would be zero so I'm not sure how to calculate it.
Well, to go the long way, you need to use the Law of Total Probability. Letting $X$ be the card you took away, and $Y$ the card you subsequently draw from the remaining deck.
$$\mathsf P(Y=\mathrm J\heartsuit) = \mathsf P(Y=\mathrm J\heartsuit\mid X\neq \mathrm J\heartsuit)~\mathsf P(X\neq \mathrm J\heartsuit)+0\cdot\mathsf P(X=\mathrm J\heartsuit)$$
Well, now, $\mathsf P(X\neq\mathrm J\heartsuit) = 51/52$ is the probability that the card taken is one from the 51 not-jack-of-hearts.
Also $\mathsf P(Y=\mathrm J\heartsuit\mid X\neq \mathrm J\heartsuit)$ is the probability that the subsequent selection is the Jack of Hearts when given that that card is among the 51 remaining cards.
Alternatively, the short way is to consider : When given a shuffled deck of 52 standard cards, what is the probability that the second from the top is the Jack of Hearts?
The probability is 1/52.
It doesn't matter that one card was moved somewhere else.