"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.
Partitioning the deck into slices is almost right idea, but you need to exclude the kings from the count. This is because only twenty of the twenty-one slices contain a king in your method, so the slices do not all have the same expected size in your counting scheme.
There are $240$ non-kings, so the expected number of non-kings in each slice is $240/21$. This means the expected number of cards dealt is $5\times (240/21)+5$, where the $+5$ accounts for the five kings which come after the first five slices.
By the way, the question asks for "maximizing your winning chances", but in your solution you compute the "expected number of cards dealt". I see no reason for these to be the same concept, and would expect that that you would actually maximize your chances by choosing the mode, not the mean.
Best Answer
Your reasoning was no so evident for me at first, so I decided to dig a little and what I first found was a table with an entry that agrees with your interpretation, but then I wanted to know where that multichoose formula comes from, and after a while I found useful this explanation using the bars and stars approach.
In your case, for example, we would represent the $10$ cards of the hand as "stars" and then we would put $52-1$ "bars" between them to represent the idea of $52$ different types$^1$ of cards. The stars that are to the left of a bar are of the same type. We have then to be careful with the extreme case where all the $10$ stars are to the left of a bar, meaning that the all the $10$ cards are all of the same type. As you and @JMoravitz correctly pointed out, this is possible because the size of the hand is not greater than the multiplicity of each type in the superdeck.
Under this approach we could imagine that we have a total of $52+10-1$ bins to place the stars and the bars, and then ask in how many ways we could place the $10$ stars out of a total $52+10-1$ available bins. Now the answer is more evident to me, in
$$\binom{52+10-1}{10}$$
ways. Once you place the stars there is only one way to put the bars, since the order does not matter.
EDIT: I was wondering why Bose-Einstein would be a hint in the problem. Following the perspective proposed in this video, it seems that this problem is analogous to the Bose-Einstein statistics in that we could thought of the $10$ cards (stars) in the hand as indistinguishable particles and the $52^*$ types of cards in a deck as energy states where those particles could condensate (don't get me wrong, I am not a quantum physicist). In any case, it's not clear how that would be a hint to solve the problem...any hint?
$^*$ Thanks to Michael's answer I understood that the number of energy states is $52$ and not $52+10-1$, as I originally wrote in my edit...although it still seems a rather sophisticated hint for me.
$^1$ Here the type of a card is its rank and suit.