Let $E \vert K$ be a field extension; to play it safe, let's assume the characteristic is $0$, and all dimensions are finite. It is useful for what follows to introduce an explicit notation for scalar restriction: If $V$ is any vector space (or later $\mathfrak g$ a Lie algebra) over $E$, let's call $R_{E \vert K} V$ (or later $R_{E\vert K} \mathfrak g$) the same additive group (Lie algebra) but viewed as a vector apace (Lie algebra) over $K$.
On the other hand one has scalar extension: If $V$ is a vector space (or $\mathfrak g$ a Lie algebra) over $K$, then $V_E := E \otimes_K V$ (or $\mathfrak g_E := E \otimes_K \mathfrak g$) is a vector space (Lie algebra) over $E$. (The Lie bracket on $\mathfrak g_E$ is defined as the unique bilinear map satisfying $[e_1 \otimes g_1, e_2 \otimes g_2] = e_1e_2 \otimes [g_1, g_2]$.)
Note that in spite of what one might think when first hearing the names, scalar restriction and scalar extension are not at all inverse to each other.
If we first look at vector spaces for a moment, everybody has learned in linear algebra that for an $E$-vector space $W$, we have that $R_{E\vert K} W$ is a $K$-vector space of dimension $[E:K] \cdot \mathrm{dim}_E(W)$, and for a $K$-vector space $V$, we have that $V_E$ is an $E$-vector space of dimension $\mathrm{dim}_E(V_E) = \mathrm{dim}_K(V)$. Consequently
$$R_{E\vert K}(V_E) \simeq V^{[E:K]} \qquad \text{ (iso of $K$-vector spaces)}$$
$$(R_{E\vert K}(W))_E \simeq W^{[E:K]} \qquad \text{ (iso of $E$-vector spaces)}$$
So now one might think that whilst not being the identity, the composition of scalar restriction with scalar extension would always come out as the $[E:K]$-fold direct sum in the respective categories. As the above shows this is the case for vector spaces. It is also true for abelian Lie algebras because they are basically the same as vector spaces. But as soon as we look at more intricate structures, it will in general not be the case. And, one runs into different problems depending on whether the scalars get "first restricted, then extended" or the other way around.
Extension followed by restriction:
You seem to think mostly about composition of restriction and extension this way around: Starting with a Lie algebra $\mathfrak g$ over the "small" field $K$ and relating its extension $\mathfrak g_E$ to some direct sum of it. Well here the first problem is that they are incomparable insofar as $\mathfrak g_E$ is an $E$-Lie algebra but $\mathfrak g$ (and any direct sum thereof) is a $K$-Lie algebra. So the only object that has a chance to be isomorphic to $\mathfrak g^{[E:K]}$ is not $\mathfrak g_E$ itself, but $R_{E\vert K} (\mathfrak g_E)$. However, in general it is not:
Example 1: Let $\mathfrak g$ be either $:= \mathfrak{sl}_2(\mathbb R)$ or the simple $\mathbb R$-Lie algebra $\mathfrak g := \mathfrak{su}_2$. Then in either case $\mathfrak g_{\mathbb C} \simeq \mathfrak{sl}_2(\mathbb C)$ and hence $R_{\mathbb C \vert \mathbb R}(\mathfrak g_{\mathbb C})$ is a simple Lie algebra, also known as $\mathfrak{so}(3,1)$, which is $\not \simeq \mathfrak g \oplus \mathfrak g$.
Example 1*: Actually, let $\mathfrak g$ be any absolutely simple Lie algebra over $K$ (that is, one such that any scalar extension of it stays simple; for example, all split simple Lie algebras are of that kind). Then by definition $\mathfrak g_E$ is simple, and one can show that this makes $R_{E\vert K}(\mathfrak g_E)$ simple (as a $K$-Lie algebra) as well, so it cannot be isomorphic to $\mathfrak g^{[E:K]}$ as soon as $[E:K] \ge 2$.
Of course if one chooses a $K$-basis $e_1, ..., e_r$ for $E$, one can identify the underlying $K$-vector space of $\mathfrak g_E$ with an $r$-fold direct sum of $\mathfrak g$, however, this direct sum is not a direct sum of Lie algebras, as discussed in the comments. This means that in general $R_{E\vert K} ( \mathfrak g_E) \not \simeq \mathfrak g^{[E:K]}$ if the RHS is understood as a direct sum of Lie algebras.
Restriction followed by extension:
In more advanced algebra, one learns that this can go wrong for various reasons when, instead of vector spaces or Lie algebras, one looks at fields.
Example 2: Let $K:=\mathbb Q, E:=\mathbb Q(\sqrt[3]{2})$. Then $$(R_{E\vert K}(E))_E = E\otimes_K E \simeq E \times L$$ where $L= E(\zeta_3) = E(\sqrt{-3})$ is the normal closure of $E$. Actually, that is an isomorphism of rings, which when viewed as isomorphism of $E$-algebras we should write
$$ (R_{E\vert K}(E))_E \simeq E \times R_{L \vert E} (L).$$
In particular $(R_{E\vert K}(E))_E \not \simeq E \times E \times E$ as one would have naively hoped for in analogy with the vector space case. Compare https://en.wikipedia.org/wiki/Tensor_product_of_fields.
Now this dampens our hope for Lie algebras (and other kinds of algebras), as indeed:
Example 3: Let $E \vert K$ as in example 2, and let $\mathfrak g := \mathfrak{sl}_n(E)$. Then
$$(R_{E\vert K}(\mathfrak{g}))_E \simeq \mathfrak g \oplus R_{L\vert E}(\mathfrak{g}_L)$$
Now funnily, that second factor throws us back to composition the other way around, first extension and then restriction, but we saw in example 1* that (although $[L:E]=2$) that factor is not $\simeq \mathfrak g \oplus \mathfrak g$ but rather a simple (six-dimensional) $E$-Lie algebra. So we also have
$$(R_{E\vert K}(\mathfrak{g}))_E \not \simeq \mathfrak g^{[E:K]}.$$
One good case for restriction followed by extension: If $E\vert K$ is Galois and $\mathfrak g$ a split simple $E$-Lie algebra, $$(R_{E\vert K}\mathfrak g)_E \simeq \mathfrak g^{[E:K]}.$$
(Example 3 showed that the condition of $E\vert K$ being Galois is necessary. I am not sure if the condition that $\mathfrak g$ is split is necessary, i.e. without it it I neither have a proof nor a counterexample to the statement.)
In any case, both conditions are automatically satisfied for $E\vert K = \mathbb C \vert \mathbb R$. That is:
If $\mathfrak g$ is a complex semisimple Lie algebra, then $(R_{\mathbb C \vert \mathbb R}(\mathfrak g))_\mathbb C \simeq \mathfrak g \oplus \mathfrak g$.
Two final notes:
a) The examples above deal with (semi)simple Lie algebras and should give an idea what to expect and not to expect there. As noted earlier, the other extreme of abelian Lie algebras is trivially well-behaved like vector spaces. I have almost no idea what happens for Lie algebras between these extremes, i.e. solvable ones.
b) The answer in Precise connection between complexification of $\mathfrak{su}(2)$, $\mathfrak{so}(1,3)$ and $\mathfrak{sl}(2, \mathbb{C})$ which you link to in the comments would greatly benefit from using a notation for scalar restriction as above. Check for yourself that interpreted correctly, nothing in there contradicts anything in here, but one really needs to be careful to interpret it correctly. E.g. the first iso in there,
$$\mathfrak{so}(1,3) \simeq \mathfrak{su}(2)_\mathbb{C}$$
with my notation should be written
$$\mathfrak{so}(1,3) \simeq R_{\mathbb C \vert \mathbb R}(\mathfrak{su}(2)_\mathbb{C})$$
to make clear it's an iso of $\mathbb R$-Lie algebras, and actually came up in my example 1. And in point 2 of that answer, they're looking at an extension of a restriction of an extension, which would make any notation awkward, but if one in a first step abbreviated $\mathfrak{su}(2)_{\mathbb C} \simeq \mathfrak{sl}_2 (\mathbb C)$ it basically says the same as my last line in the "good case" applied to $\mathfrak g = \mathfrak{sl}_2(\mathbb C)$.
Added in response to comments: Indeed I stress that if $V = \mathbb C^2$, then $Res_{\mathbb C\vert \mathbb R}V \simeq \mathbb R^4$.
While the nomenclature might be a little misleading at first, this is indeed the standard meaning of scalar restriction. But of course the thing you have in mind is also studied. What you want is:
Given a complex Lie algebra $\mathfrak G$, is there a real Lie algebra $\mathfrak g$ such that $\mathfrak g_{\mathbb C} \simeq \mathfrak G$?
Such a $\mathfrak g$ is called a real form of $\mathfrak G$, and finding such a real form would be an actual "inverse" of scalar extension (compare my answer here, especially beginning of second paragraph, where I also note that we want this, but it's not what is called scalar restriction).
Now again, if we were just interested in vector spaces, this is trivial: If $V$ is an $n$-dimensional vector space over $E$, then $K^n$ is a $K$-form of $V$, and it is the unique one up to iso of $K$-vector spaces. But again, as soon as we are interested in more intricate structures, we face dire problems:
Problem A: Not every complex Lie algebra has a real form.
See example 1.36 here.
Widely celebrated result avoiding problem A for a large class of objects: Every semisimple complex Lie algebra has a real form. Actually, it even has a form over $\mathbb Q$; or even over $\mathbb Z$. This is widely credited to Chevalley and more or less the content of volume 8 of Bourbaki's treatise on Lie Groups and Algebras.
Sounds good. But:
Problem B: Every non-zero complex semisimple Lie algebra has more than one real form. I.e. there are non-isomorphic real Lie algebras which have isomorphic complexifications. E.g. $\mathfrak{su}_2 \not \simeq \mathfrak{sl}_2(\mathbb R)$, but both have complexification $\simeq \mathfrak{sl}_2(\mathbb C)$.
Actually, it is a celebrated result of E. Cartan's to classify all real semisimple Lie algebras, i.e. real forms of complex Lie algebras. I wrote my thesis partly about general approaches of this classification, partly about the exact classification over $p$-adic fields. Results over general characteristic $0$ fields are due largely to Tits in the 1960s and 1970s; I think that to this day some questions are open. Compare https://math.stackexchange.com/a/3677910/96384, Classification of real semisimple lie algebras or this recent question just about possible forms of $\mathfrak{sl}_2$: Lie algebra $\mathfrak{sl}_2 \mathbb{C}$ has only these two real forms $\mathfrak{sl}_2 \mathbb{R}$ and $\mathfrak{su}_2$?
