For a complex manifold $N$, denote by $H^0(N,TN)$ the space of holomorphic vector fields on $N$. Let $M_i, i=1,2$ be compact complex manifolds, $M=M_1\times M_2$ their product. Prove that $$H^0(M,TM) \cong H^0(M_1,TM_1) \oplus H^0(M_2,TM_2).$$ Roughly every vector field on $M$ is uniquely the sum of a vector field on $M_1$ and a vector field on $M_2$. Give examples to show that if compactness is dropped then the result may or may not hold.
So I was thinking this follows from $\Gamma$ or $H^0$ preserving direct sums. Since $$TM=T(M_1\times M_2) \cong TM_1\oplus TM_2$$ applying $H^0$ we have that $$H^0(M,TM) \cong H^0(M_1, TM_1) \oplus H^0(M_2,TM_2)$$ but the latter part of the problem makes me wonder why this wouldn't work unless $M$ is compact? The proof for $TM \cong TM_1\oplus TM_2$ doesn't rely on compactness and I don't think the application of the cohomology or global section functor does neither?
So this isn't partially-answered-in-the-comments, here's a detailed sketch.
First, a diagram in the style of Kobayashi:
For $k = 1$, $2$, let $\pi_{k}:M \to M_{k}$ denote the projection map. There is an isomorphism of holomorphic vector bundles over $M$, $$ T(M_{1} \times M_{2}) = \pi_{1}^{*}TM_{1} \oplus \pi_{2}^{*} TM_{2}. $$ A holomorphic vector field $X$ on the product, a.k.a. a holomorphic section of the left-hand bundle, may therefore be identified with an ordered pair $(X_{1}, X_{2})$ of holomorphic sections of the respective pullback summands on the right.
A priori, each component $X_{k}$ is defined on the product $M$ and takes values in the pullback bundle $\pi_{k}^{*}TM_{k}$. It suffices to show $X_{k}$ may be identified with a holomorphic vector field on $M_{k}$; loosely, that "each section is constant in the other variable."
Generally (i.e., without compactness), the pullback $\pi_{1}^{*}TM_{1}$ is holomorphically trivial when restricted to a slice $\{p\} \times M_{2}$ (fibers shown in gray); the total space is $M_{2} \times T_{p}M_{1}$, equipped with projection to $M_{2}$. A section of $\pi_{1}^{*}TM_{1}$, restricted to $\{p\} \times M_{2}$, may therefore be viewed as a holomorphic mapping from $M_{2}$ into the complex vector space $T_{p}M_{1}$.
Because $M_{2}$ is compact, every holomorphic map from $M_{2}$ to $T_{p}M_{1}$ is constant. (This is the point where compactness and "holomorphic" are essential.) Particularly, $X_{1}$ restricted to $\{p\} \times M_{2}$ is constant.
On the other hand, for each point $q$ in $M_{2}$, the restriction of $\pi_{1}^{*}TM_{1}$ to $M_{1} \times \{q\}$ is isomorphic to $TM_{1}$ as a holomorphic vector bundle. Particularly, $X_{1}$ restricted to $M_{1} \times \{q\}$ may be identified with a holomorphic vector field on $M_{1}$.
Finally, since the section $X_{1}$ on $M$ is "constant in $q$," it may be identified with a section of $TM_{1}$.
The same argument with the roles of the factors swapped shows $X_{2}$ may be identified with a section of $TM_{2}$.
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To give a non-compact manifold where the same decomposition holds, it suffices to find a non-compact manifold on which every holomorphic function is constant.
The comment of user10354138 gives an example where the stated decomposition of vector fields does not hold.