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So I just proved the following proposition:

Let $M_1, \dots, M_k$ be smooth manifolds, and for each $j$, let $\pi_j: M_1 \times \cdots \times M_k \rightarrow M_j$ be the projection onto the $M_j$ factor. For any point $p = (p_1, \dots, p_k) \in M_1 \times \cdots \times M_k$, the map $\alpha: T_p(M_1 \times \cdots \times M_k) \rightarrow T_{p_1}M_1 \oplus \cdots \oplus T_{p_k}M_k$ defined by $\alpha(v) = \big(d(\pi_1)_p(v), \dots, d(\pi_k)_p(v) \big)$ is an isomorphism.

Now, I proved $\alpha$ was bijective by using local coordinates (an arbitrary set of local coordinates), but I felt uneasy doing this. I am confident my proof is correct, but the entire time I was doing it, I was always double-checking that whatever statement didn't depend on local coordinates. In my study of smooth manifold theory, is it a good idea to avoid using local coordinates? I don't want to develop bad habits.

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    I think it is a fairly standard technique to use local coordinates and then show that whatever you proved is independent of your choice of coordinates. I'm not sure if this is always the fanciest way to do things, but many times it is the most straightforward (I'm mostly a physicist though so may not be the best to ask about bad habits in smooth manifold theory). – 2017-02-01

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It's alright to prove it using local coordinates but in this case, you can also do it without introducing local coordinates at all. A tangent vector at $p = (p_1,\dots,p_k)$ is an equivalence class of curves $c \colon I \rightarrow M_1 \times \dots \times M_k$ with $c(0) = p$. For each such curve, we can write

$$ c(t) = (c_1(t), \dots, c_k(t)) $$

where $c_i(t) = \pi_i \circ c$ are the components of $c$. Using this notation, the map $\alpha$ is described by

$$ \alpha([c]) = ([c_1], \dots, [c_k]) = ([\pi_1 \circ c], \dots, [\pi_k \circ c]). $$

This map is linear (this is evident from your description) and onto (given $c_i \colon I \rightarrow M_i$ we can define $c$ by $c(t) = (c_1(t), \dots, c_k(t))$ and then $\alpha([c]) = ([c_1], \dots, [c_k])$) and so by dimensional reasons it is also one-to-one and hence an isomorphism.