I want to show $$\mathbb{R}^2\otimes \mathbb{C}\cong \mathbb{C}^2.$$
My guess is to consider the map $f$ such that $(x,y)\otimes z\mapsto ((x,y),z)=(x+iy,z)=(\tilde{z},z)$ which is bijective. But I can't show that it preserves addition because I don't know how to make sense of $f((x,y)\otimes z+(x,y)'\otimes z')$. Do we get linearity automatically from the universal property ? I am new to tensor products, so I am a bit confused about how things exactly work.
Here's a general thing about tensor product, if you want to define linear map $f\colon A\otimes B\to C$, you first need to define bilinear map $g\colon A\times B\to C$, i.e. map that is linear in each argument separately. Then the universal property of tensor product guarantees that there is a unique linear map $f\colon A\otimes B\to C$ such that $f(a\otimes b) = g(a,b)$.
I'm not even sure what exactly is the map that you are trying to define, because equalities $((x,y),z) = (x+iy,z) = (\bar z,z)$ make zero sense, none of it is right.
The relevant general construction here is called extension of scalars. This works for modules, but in this context, we have some vector space $V$ over field $k$ and field $K\supseteq k$, and we want to define vector space $V^K$ over $K$ such that it extends addition and multiplication by scalars from $k$ on $V$. We do that by tensoring: $V^K:=K\otimes_k V$ where we define multiplication by scalars from $K$ by formula $\alpha\cdot(\beta\otimes v):=(\alpha\beta)\otimes v$. We can identify vectors from $V$ with vectors from $V^K$ of the form $1\otimes v$, for $v\in V$, and this defines $k$-linear embedding $v\mapsto 1\otimes v$ since for $\alpha \in k$ and $v\in V$ we have $\alpha v \mapsto 1\otimes \alpha v = \alpha\otimes v = \alpha\cdot(1\otimes v)$.
I've included this general construction since it should give you the correct idea what the factor $\mathbb C$ in $\mathbb R^2\otimes_{\mathbb R} \mathbb C$ does - it defines multiplication by complex numbers on $\mathbb R^2$. Therefore, any map $\mathbb R^2\otimes_{\mathbb R} \mathbb C\to \mathbb C^2$ should take that in account, explicitly $(x,y)\otimes z$ should be mapped to $z\cdot (x,y)$ where $\cdot$ is multiplication by scalars in $\mathbb C^2$, i.e. $z\cdot (x,y) = (zx,zy)$.
So, we define $\mathbb R$-bilinear map $\mathbb R^2\times \mathbb C\to \mathbb C^2$ by the formula $((x,y),z) \mapsto (zx,zy)$. You can verify that $$(\alpha_1(x_1,y_1)+\alpha_2(x_2,y_2),z)\mapsto \alpha_1(zx_1,zy_1)+\alpha_2(zx_2,zy_2)$$ and $$((x,y),\alpha_1z_1+\alpha_2z_2)\mapsto \alpha_1(z_1x,z_1y)+\alpha_2(z_2x,z_2y)$$ for all real $\alpha_1,\alpha_2,x,y,x_1,x_2,y_1,y_2$ and complex $z,z_1,z_2$.
By the universal property of tensor product, there is a unique $\mathbb R$-linear map $\mathbb R^2\otimes_{\mathbb R}\mathbb C\to \mathbb C^2$ that maps $(x,y)\otimes z$ to $(zx,zy)$. We still need to verify that the map is $\mathbb C$-linear, i.e. $z_1\cdot ((x,y)\otimes z_2)\mapsto z_1(z_2x,z_2y)$ but this is clear since $$z_1\cdot((x,y)\otimes z_2) = (x,y)\otimes (z_1z_2) \mapsto (z_1z_2x,z_1z_2y) = z_1(z_2x,z_2y).$$
What is left for you to do is to prove that this map is isomorphism.