Is the sum of two prime ideals in different polynomial rings, K[X_i] and K[Y_i] a prime ideal in K[X_i Y_i]?

We have $K\left[x_1,...,x_m,y_1,...,y_n\right] \cong K\left[x_1,...,x_m\right] \otimes K\left[y_1,...,y_n\right]$ (where all tensor products are over $K$), and under this isomorphism, the ideal of $K\left[x_1,...,x_m,y_1,...,y_n\right]$ generated by $P+Q$ corresponds to $P\otimes K\left[y_1,...,y_n\right] + K\left[x_1,...,x_m\right]\otimes Q$. So your question is a particular case of the following one:

Let $K$ be a field, and $A$ and $B$ be two $K$-algebras, and $P$ and $Q$ be two prime ideals of $A$ and $B$, respectively. When can we conclude that $P\otimes B+A\otimes Q$ is a prime ideal of $A\otimes B$ ?

Since $P\otimes B+A\otimes Q$ is the kernel of the canonical map $A\otimes B\to \left(A/P\right)\otimes \left(B/Q\right)$ (this is a fact of linear algebra, and I can give a proof if needed), it is clear that $P\otimes B+A\otimes Q$ is a prime ideal of $A\otimes B$ if and only if $\left(A/P\right)\otimes \left(B/Q\right)$ is an integral domain. On the other hand, we know that $A/P$ and $B/Q$ are integral domains (since $P$ and $Q$ are prime ideals). So our question becomes: When is the tensor product of two integral domains (which both are $K$-algebras) again an integral domain?

This is always true when $K$ is algebraically closed and one of the two integral domains is a finitely-generated $K$-algebra (see Proposition 4.15 (b) in J. S. Milne, Algebraic Geometry, Version 5.21). For $K$ not algebraically closed, the answer is "no" (not generally), even in your case as we can easily see (in fact, every finitely generated $K$-algebra which is an integral domain can be considered as a quotient of a polynomial ring $K\left[z_1,z_2,...,z_k\right]$ for some $k$ modulo some prime ideal, so the problems are equivalent). For a particularly simple counterexample to your claim, take $K=\mathbb R$, $n=1$, $m=1$, $P=\left(x_1^2+1\right)$ and $Q=\left(y_1^2+1\right)$ (in this case, $P\otimes Q\cong \mathbb C\otimes_{\mathbb R} \mathbb C$, and this is a commutative $\mathbb R$-algebra of dimension $4$, so it is clearly not an integral domain).

The geometric analogue to darij grinberg's answer is that you are effectively asking whether the product of two particular irreducible algebraic varieties is also irreducible. Here is how the translation works.

The two polynomial rings $K[x_1, \dots, x_m]$ and $K[y_1, \dots, y_n]$ are the coordinate rings of the affine spaces $\mathbb{A}^m_K$ and $\mathbb{A}^n_K$, and their prime ideals $P$ and $Q$ are, respectively, irreducible subvarieties $X$ and $Y$ in these. The combined ring $K[x_1, \dots, x_m, y_1, \dots, y_n]$ is the ring of $\mathbb{A}^{n + m}_K$, which is isomorphic to the product $\mathbb{A}^n_K \times_K \mathbb{A}^m_K$. Considering $P$ and $Q$ as prime ideals in the combined ring is the same as forming the varieties $X \times_K \mathbb{A}^m_K$ and $\mathbb{A}^n_K \times_K Y$, and their intersection is itself isomorphic to $X \times_K Y$. You want to know whether this is necessarily irreducible.

Over an algebraically closed field the answer is "yes". This formally follows from the theorem darij cited, but you can also understand it geometrically in that if, for example, $K = \mathbb{C}$, you can cut the ${}_K$ from the product and convince yourself that since the slices of $X \times Y$ over $Y$ are irreducible (isomorphic to $X$) and of equal dimension, the product itself is also irreducible. (To understand the second condition, think about the reducible variety $\{0\} \times \mathbb{A}^1 \cup \mathbb{A}^1 \times \{0\}$ and its map to $\mathbb{A}^1$ in, say, the second factor.)

Over a non-closed field you have darij's example. It's hard to give a totally geometric picture for this sort of thing, since geometry takes place over a closed field, but you can think about it in a sort of quasi-pictorial way. Namely, the real affine line $\mathbb{A}^1_\mathbb{R}$ (with coordinate ring $\mathbb{R}[x]$) has two kinds of points: those in $\mathbb{R}$, corresponding to prime ideals $(x - a)$, and conjugate pairs of those in $\mathbb{C}$, corresponding to prime ideals $(x^2 + 2bx + c)$ with $b^2 < c$. This means that $\mathbb{A}^1_\mathbb{R}$ is like $\mathbb{A}^1_\mathbb{C}$ with conjugate pairs of points being grouped together into one point.

In $\mathbb{A}^2_\mathbb{R}$, things are more complicated because of dimension, but for closed points, it's similar: each coordinate is like a pair of conjugate complex numbers grouped together.

Suppose you have points like $\{z, \bar{z}\}$ and $\{w, \bar{w}\}$ in $\mathbb{A}^1_\mathbb{R}$. Their product, taken in the naive sense, is going to look like four pairs: $$ \{ (z, w), (\bar{z}, \bar{w}), (z, \bar{w}), (\bar{z}, w) \}. $$ The first two and the last two are conjugate pairs, but (in the absence of coincidences like one of them being real) not conjugate to each other, so you will get two distinct points in $\mathbb{A}^2_\mathbb{R}$, which is not an irreducible set.

This corresponds to the fact that, as darij computed, the coordinate ring of this product is $\mathbb{C} \otimes_\mathbb{R} \mathbb{C}$, which is isomorphic to $\mathbb{C} \oplus \mathbb{C}$ since $\mathbb{C}$ is Galois of degree 2 over $\mathbb{R}$. And in fact, everything I've said about conjugates in the last few paragraphs has used this fact implicitly.