"all cards of the same kind?" No. A hand of cards is just a subset of the cards. Here, since we have duplicate cards we need to be a bit more careful with how we phrase it... it is a submultiset of the cards. Order within the hand does not matter. Two hands are considered the "same" if what cards in what quantities are the same in one hand compared to another. For example, a hand that contains five $A\spadesuit$ four $K\heartsuit$ and one $K\clubsuit$ is considered the same as any other hand which contains those same type of cards in the same quantities, regardless order and what deck. Here it is assumed that which deck a card comes from is irrelevant. If every deck of cards had different art styles and a different cardback and could be distinguished it would have been a very different problem, one with an answer of $\binom{520}{10}$
"What if the question had 11 decks" No, the answer will still be $\binom{10+52-1}{52-1}=\binom{10+52-1}{10}$, we did not need an $11$ there. The $10$ that appeared in the formula here happened to be our handsize not the number of decks. That the number of decks was ten is only relevant because it is greater than our number of cards in hand, allowing us to treat this as though it were pulling with replacement. If it were five decks instead, we would need to be more careful, likely needing to combine stars-and-bars with inclusion-exclusion.
"how are there $52+10-1$ slots instead of $52\cdot 10=520$ slots for ten decks of cards?" I don't know what you mean by slots here, but this is again a standard application of stars-and-bars. We have different categories. We want some collection of some number of those categories, repetition allowed order not relevant. In order to split the categories up, we only need a barrier between the categories.
Consider this small example of having five identical cookies and we ask how we can split them up between me and you. I say, we can draw a line. All cookies on the left side of the line are mine and all cookies on the right side of the line are yours. We have the possibilities: $\mid \circ \circ \circ \circ\circ$ where I get none and you get five, we have $\circ \mid \circ \circ \circ \circ$ where I get one and you get four... on up to $\circ \circ \circ \circ \circ \mid$ where I get all five and you get none. I only needed one barrier to separate these into two piles, not two barriers. Similarly, if I wanted to split it into $k$ piles I only need $k-1$ barriers.
The essence of stars-and-bars is again to distribute $n$ identical balls (cards in our hand) into $k$ distinct bins (cardtypes) where order in which balls appear in bins doesn't matter this is like picking some arrangement of $n$ stars and $k-1$ bars which can be done in $\binom{n+k-1}{k-1}$ ways, equivalently $\binom{n+k-1}{n}$ ways.
